Exercise 0.1: Computing a limit at a point: twelve direct calculations

For each line, compute the requested limit without L'Hôpital's rule and give its value, or say whether it is $+\infty$ , $-\infty$ or does not exist.

One method for all twelve lines: substitute first. If you get a number, that is the limit. If you get $\frac{0}{0}$ , resolve the indeterminate form with the technique it calls for (factor, multiply by the conjugate, combine over a common denominator, bring out $\frac{\sin u}{u}$ ), then substitute again. If you get $\frac{k}{0}$ with $k\neq 0$ , study the sign of the denominator on each side.

a) $\lim_{x\to -2}\frac{x^{2}-3x}{x+4}$ .
b) $\lim_{x\to -1}\frac{x^{2}+6x+5}{x^{2}-x-2}$ .
c) $\lim_{x\to -2}\frac{x^{3}+8}{x+2}$ .
d) $\lim_{x\to 5}\frac{\sqrt{x+4}-3}{x-5}$ .
e) $\lim_{x\to 7}\frac{x-7}{\sqrt{x+2}-3}$ .
f) $\lim_{x\to 2}\frac{\frac{2}{x}-1}{x-2}$ .
g) $\lim_{h\to 0}\frac{(-1+h)^{3}+1}{h}$ .
h) $\lim_{x\to 0}\frac{\tan 6x}{\sin 4x}$ .
i) $\lim_{x\to 0}\frac{1-\cos 2x}{x\sin x}$ .
j) $\lim_{x\to -2^{-}}\frac{x-1}{x+2}$ and $\lim_{x\to -2^{+}}\frac{x-1}{x+2}$ .
k) $\lim_{x\to -1}\frac{x-4}{(x+1)^{2}}$ .
l) $f(x)=x^{2}+1$ if $x<1$ and $f(x)=4-x$ if $x\geq 1$ : $\lim_{x\to 1^{-}}f(x)$ , then $\lim_{x\to 1}f(x)$ .

Type your answers, the page tells you right or wrong 0/14

a)

The limit

b)

The limit

c)

The limit

d)

The limit

e)

The limit

f)

The limit

g)

The limit

h)

The limit

i)

The limit

j)

Limit at

-2^{-}

Limit at

-2^{+}

k)

The limit

l)

Limit at

1^{-}

\lim_{x\to 1}f(x)

Show the solution

Answers

a) Direct substitution; $5$
b) $\frac{x+5}{x-2}$ ; $-\frac{4}{3}$
c) $x^{2}-2x+4$ ; $12$
d) $\frac{1}{\sqrt{x+4}+3}$ ; $\frac{1}{6}$
e) $\sqrt{x+2}+3$ ; $6$
f) $-\frac{1}{x}$ ; $-\frac{1}{2}$
g) $3-3h+h^{2}$ ; $3$
h) $\frac{6}{4}$ by the fundamental limits; $\frac{3}{2}$
i) $2\cdot\frac{\sin x}{x}$ ; $2$
j) $\frac{-3}{0^{\mp}}$ ; $+\infty$ from the left, $-\infty$ from the right
k) $\frac{-5}{0^{+}}$ ; $-\infty$
l) $2$ from the left, $3$ from the right; the limit does not exist

The move that decides everything: substitute BEFORE manipulating. The form you get chooses the technique: a number is read off, $\frac{0}{0}$ hides a common factor to bring out, $\frac{k}{0}$ signals an infinite limit whose sign is all that remains to find.

a)

The denominator equals $2\neq 0$ : direct substitution, $\frac{4+6}{2}=5$ .

b)

Form $\frac{0}{0}$ , so $x+1$ is a factor of both top and bottom: $\frac{(x+1)(x+5)}{(x+1)(x-2)}=\frac{x+5}{x-2}\to\frac{4}{-3}=-\frac{4}{3}\approx-1.333$ .

c)

Form $\frac{0}{0}$ . Sum of cubes: $x^{3}+8=(x+2)\left(x^{2}-2x+4\right)$ , so the limit is $4+4+4=12$ . Pitfall: writing $x^{2}+2x+4$ ; the middle sign is the OPPOSITE of the one in the binomial.

d)

Form $\frac{0}{0}$ . Conjugate: $\frac{(x+4)-9}{(x-5)\left(\sqrt{x+4}+3\right)}=\frac{1}{\sqrt{x+4}+3}\to\frac{1}{6}\approx 0.167$ .

e)

Form $\frac{0}{0}$ , with the root in the denominator: multiply by $\sqrt{x+2}+3$ , and $\frac{(x-7)\left(\sqrt{x+2}+3\right)}{(x+2)-9}=\sqrt{x+2}+3\to 6$ .

f)

Form $\frac{0}{0}$ . Common denominator on top: $\frac{2}{x}-1=\frac{2-x}{x}$ , so the expression equals $\frac{2-x}{x(x-2)}=-\frac{1}{x}\to-\frac{1}{2}$ . Pitfall: $2-x=-(x-2)$ , and the minus sign is easily lost.

g)

Form $\frac{0}{0}$ . Expand: $(-1+h)^{3}=-1+3h-3h^{2}+h^{3}$ , so the quotient equals $3-3h+h^{2}\to 3$ .

h)

Form $\frac{0}{0}$ . Bring out the fundamental limits: $\frac{\tan 6x}{6x}\cdot\frac{4x}{\sin 4x}\cdot\frac{6}{4}\to 1\cdot 1\cdot\frac{3}{2}=\frac{3}{2}$ , since $\frac{\tan u}{u}=\frac{\sin u}{u}\cdot\frac{1}{\cos u}\to 1$ .

i)

Form $\frac{0}{0}$ . Double-angle identity: $1-\cos 2x=2\sin^{2}x$ , so the expression equals $\frac{2\sin^{2}x}{x\sin x}=2\cdot\frac{\sin x}{x}\to 2$ .

j)

Form $\frac{-3}{0}$ : an infinite limit. To the left of $-2$ , $x+2<0$ , so $\frac{-3}{0^{-}}\to+\infty$ ; to the right, $x+2>0$ , so $\frac{-3}{0^{+}}\to-\infty$ . Pitfall: forgetting that the numerator is NEGATIVE, which reverses both signs.

k)

Form $\frac{-5}{0}$ : the square $(x+1)^{2}$ is positive on both sides, so $\frac{-5}{0^{+}}\to-\infty$ from the left and from the right alike, and the limit is $-\infty$ .

l)

From the left, $f(x)=x^{2}+1\to 2$ ; from the right, $f(x)=4-x\to 3$ . The two one-sided limits differ, so $\lim_{x\to 1}f(x)$ does not exist, even though $f(1)=3$ exists.

Check: on a calculator, evaluate the expression at $a-0.001$ and at $a+0.001$ . The two values should bracket the limit you found; for a form $\frac{k}{0}$ , they should be very large, with the predicted sign on each side.

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Exercise 0.2: Computing a limit at infinity: ten direct calculations

For each line, compute the indicated limit at $+\infty$ or at $-\infty$ without L'Hôpital's rule and give its value, or say whether it is $+\infty$ or $-\infty$ .

One method for all ten lines: spot the DOMINANT term in the numerator and in the denominator and factor it out, remembering that $\sqrt{x^{2}}=\left|x\right|$ , which equals $-x$ at $-\infty$ ; resolve a form $\infty-\infty$ with a square root by the conjugate; for the exponential, the arctangent and the logarithm, fall back on the behaviour of the basic function.

a) $\lim_{x\to-\infty}\left(3x^{4}-x^{5}+2\right)$ .
b) $\lim_{x\to-\infty}\frac{6x^{3}-2x}{1-4x^{3}}$ .
c) $\lim_{x\to-\infty}\frac{2x^{3}+1}{x^{2}-4x}$ .
d) $\lim_{x\to-\infty}\frac{\sqrt{16x^{2}+x}}{1-2x}$ .
e) $\lim_{x\to+\infty}\left(\sqrt{4x^{2}+5x}-2x\right)$ .
f) $\lim_{x\to+\infty}\frac{3+\sin 2x}{x+1}$ .
g) $\lim_{x\to+\infty}\frac{6\cdot 3^{x}+1}{2\cdot 3^{x}-2^{x}}$ .
h) $\lim_{x\to-\infty}\arctan\left(x^{2}\right)$ .
i) $\lim_{x\to+\infty}\left[\ln\left(2x^{2}+1\right)-2\ln(x+5)\right]$ .
j) $\lim_{x\to+\infty}\frac{5x^{\frac{3}{2}}-x}{2x\sqrt{x}+7}$ .

Type your answers, the page tells you right or wrong 0/10

a)

The limit

b)

The limit

c)

The limit

d)

The limit

e)

The limit

f)

The limit

g)

The limit

h)

The limit

i)

The limit

j)

The limit

Show the solution

Answers

a) Like $-x^{5}$ ; $+\infty$
b) $\frac{6}{-4}$ ; $-\frac{3}{2}$
c) Like $2x$ ; $-\infty$
d) $\frac{-4}{-2}$ , with $\sqrt{x^{2}}=-x$ ; $2$
e) $\frac{5x}{\sqrt{4x^{2}+5x}+2x}$ ; $\frac{5}{4}$
f) Squeezed between $\frac{2}{x+1}$ and $\frac{4}{x+1}$ ; $0$
g) $\frac{6}{2}$ ; $3$
h) $\arctan u$ , $u\to+\infty$ ; $\frac{\pi}{2}\approx 1.571$
i) $\ln\frac{2x^{2}+1}{(x+5)^{2}}$ ; $\ln 2\approx 0.693$
j) $\frac{5}{2}$ ; $\frac{5}{2}$

The move that decides everything: factor out the dominant term, and only that term. Everything else tends to $0$ or to a constant; all that is left is to read a ratio of coefficients, or a sign.

a)

$3x^{4}-x^{5}+2=x^{5}\left(\frac{3}{x}-1+\frac{2}{x^{5}}\right)$ : the bracket tends to $-1$ and $x^{5}\to-\infty$ , so the limit is $+\infty$ . Pitfall: reading the term $3x^{4}$ , which is positive, when it is $-x^{5}$ that dominates.

b)

Equal degrees: divide by $x^{3}$ , $\frac{6-\frac{2}{x^{2}}}{\frac{1}{x^{3}}-4}\to\frac{6}{-4}=-\frac{3}{2}$ , at $-\infty$ and at $+\infty$ alike.

c)

Higher degree on top: $\frac{2x^{3}+1}{x^{2}-4x}=\frac{x^{3}\left(2+\frac{1}{x^{3}}\right)}{x^{2}\left(1-\frac{4}{x}\right)}$ , which behaves like $2x$ , so it tends to $-\infty$ .

d)

$\sqrt{16x^{2}+x}=\left|x\right|\sqrt{16+\frac{1}{x}}=-x\sqrt{16+\frac{1}{x}}$ for $x<0$ . So the expression equals $\frac{-x\sqrt{16+\frac{1}{x}}}{x\left(\frac{1}{x}-2\right)}=\frac{-\sqrt{16+\frac{1}{x}}}{\frac{1}{x}-2}\to\frac{-4}{-2}=2$ . Pitfall: writing $\sqrt{x^{2}}=x$ , which gives $-2$ .

e)

Form $\infty-\infty$ . Conjugate: $\frac{\left(4x^{2}+5x\right)-4x^{2}}{\sqrt{4x^{2}+5x}+2x}=\frac{5x}{x\left(\sqrt{4+\frac{5}{x}}+2\right)}\to\frac{5}{4}$ . Pitfall: concluding $0$ because the two terms “look alike”.

f)

The sine is bounded: $2\leq 3+\sin 2x\leq 4$ , so $\frac{2}{x+1}\leq\frac{3+\sin 2x}{x+1}\leq\frac{4}{x+1}$ for $x>-1$ , and both bounds tend to $0$ . By the Squeeze Theorem, the limit is $0$ .

g)

The dominant term is $3^{x}$ in both the numerator and the denominator: $\frac{6+3^{-x}}{2-\left(\frac{2}{3}\right)^{x}}\to\frac{6}{2}=3$ , since $\left(\frac{2}{3}\right)^{x}\to 0$ .

h)

At $-\infty$ , $x^{2}\to+\infty$ , and $\arctan u\to\frac{\pi}{2}$ as $u\to+\infty$ : the limit is $\frac{\pi}{2}\approx 1.571$ . Pitfall: answering $-\frac{\pi}{2}$ by looking at $x$ instead of $x^{2}$ .

i)

Form $\infty-\infty$ . Combine the logarithms: $\ln\frac{2x^{2}+1}{(x+5)^{2}}$ , and the quotient tends to $2$ (equal degrees), so by continuity of the logarithm the limit is $\ln 2\approx 0.693$ .

j)

$x\sqrt{x}=x^{\frac{3}{2}}$ : the dominant term is $x^{\frac{3}{2}}$ in both the numerator and the denominator, since $x^{\frac{3}{2}}$ outgrows $x$ . Divide: $\frac{5-x^{-\frac{1}{2}}}{2+7x^{-\frac{3}{2}}}\to\frac{5}{2}$ .

Check: on a calculator, evaluate the expression at $x=10^{4}$ (or $-10^{4}$ ), then at $x=10^{6}$ ; the values should get closer to the limit you found, or keep growing with the predicted sign.

Exercise 1: Reading limits from a graph

The figure shows a function $f$ with a JUMP at $x=1$ and a HOLE at $x=3$ . Solid dots are values the function takes; open circles are excluded values.

a) Find $\lim_{x\to 1^{-}}f(x)$ , $\lim_{x\to 1^{+}}f(x)$ , $\lim_{x\to 1}f(x)$ and $f(1)$ . Comment on your results.
b) Find $\lim_{x\to 3^{-}}f(x)$ , $\lim_{x\to 3^{+}}f(x)$ , $\lim_{x\to 3}f(x)$ and $f(3)$ . How does the situation differ from the one at $x=1$ ?
c) Is the function continuous at $x=1$ ? At $x=3$ ? Justify with the three conditions for continuity, and name the type of discontinuity in each case.
d) Can $f$ be REDEFINED at a single point to make it continuous at $x=3$ ? And at $x=1$ ? Explain what this difference reveals about the deep nature of the two discontinuities.

Type your answers, the page tells you right or wrong 0/13

a)

\lim_{x\to 1^{-}}f(x)

\lim_{x\to 1^{+}}f(x)

\lim_{x\to 1}f(x)

f(1)

b)

\lim_{x\to 3^{-}}f(x)

\lim_{x\to 3^{+}}f(x)

\lim_{x\to 3}f(x)

f(3)

c)

At

x=1

At

x=3

d)

Removable at

x=3

?

Value to assign at

3

Removable at

x=1

?

Show the solution

Answers

a) $2$ and $3$ : no limit; $f(1)=3$
b) Limit $1$ , but $f(3)=2$
c) Jump at $1$ , hole at $3$
d) Set $f(3)=1$ ; nothing works at $1$

a)

Approaching from the LEFT, follow the rising branch that reaches the open circle at height 2: $\lim_{x\to 1^{-}}f(x)=2$ . From the RIGHT, follow the falling branch that starts at height 3: $\lim_{x\to 1^{+}}f(x)=3$ . The two one-sided limits DIFFER, so $\lim_{x\to 1}f(x)$ DOES NOT EXIST. Finally, the solid dot gives $f(1)=3$ : the function is indeed defined at 1, even though its limit does not exist. Being defined and having a limit are two independent questions.

b)

From the left, the falling branch tends to the open circle at height 1: $\lim_{x\to 3^{-}}f(x)=1$ . From the right, the rising branch also starts at height 1: $\lim_{x\to 3^{+}}f(x)=1$ . The two agree, so $\lim_{x\to 3}f(x)=1$ : the limit EXISTS. But the solid dot is at height 2: $f(3)=2\ne 1$ . The reverse of the situation at $x=1$ : there the limit was missing; here it exists, and it is the value of the function that is in the wrong place.

c)

At $x=1$ : $f(1)=3$ exists, but the limit does not; the second condition fails, and $f$ is DISCONTINUOUS, with a JUMP (the two one-sided limits are finite but different). At $x=3$ : $f(3)=2$ exists, the limit exists and equals 1, but $1\ne 2$ ; the third condition fails, and $f$ has a HOLE (a removable discontinuity).

d)

At $x=3$ : YES. It is enough to set $f(3)=1$ , the value of the limit: the solid dot drops down to fill the hole and the curve becomes continuous. That is why this is called a REMOVABLE discontinuity. At $x=1$ : NO, no value can work: the curve arrives at height 2 from the left and leaves from height 3 on the right, and a single point cannot be at two heights at once. The deep difference: the hole is an accident of DEFINITION (the function knows where it wants to go, it was just recorded wrongly at one point), while the jump is an accident of BEHAVIOUR (the function itself tears apart), and no change at a single point can mend a tear.

Exercise 2: Resolving indeterminate forms

An indeterminate form $\frac{0}{0}$ is not a dead end: it is the sign that a common factor is hiding, waiting to be cancelled.

a) Compute $\lim_{x\to 2}\frac{x^{2}-4}{x-2}$ by factoring. Why are we ALLOWED to cancel $x-2$ , when this very factor vanishes at 2?
b) Compute $\lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}$ using the conjugate.
c) Compute $\lim_{x\to 1}\frac{x^{3}-1}{x^{2}-1}$ , then $\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x^{2}+x}\right)$ (indeterminate form $\infty-\infty$ ).
d) Compute $\lim_{x\to 3}\frac{x-3}{|x-3|}$ , treating the two sides separately. What should you conclude, and what general lesson does the absolute value teach?

Show the solution

Answers

a) $4$
b) $\frac{1}{6}$
c) $\frac{3}{2}$ , then $1$
d) $-1$ and $+1$ : the limit does not exist

a)

$\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$ for all $x\ne 2$ , hence the limit $2+2=4$ . The right to cancel comes from the very definition of the limit: it looks ONLY at the $x$ near 2, never at $x=2$ itself. For those $x\ne 2$ , the quotient is exactly equal to $x+2$ , so the two functions have the same limit. We never divide by zero; we simply set aside a point the limit does not care about.

b)

Multiply top and bottom by the conjugate $\sqrt{x+9}+3$ : $\frac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{x}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}$ . The offending factor $x$ is gone, and the limit is $\frac{1}{3+3}=\frac{1}{6}$ . That is exactly what the conjugate is for: turning a difference of square roots into a difference of squares, which simplifies.

c)

$\frac{x^{3}-1}{x^{2}-1}=\frac{(x-1)(x^{2}+x+1)}{(x-1)(x+1)}=\frac{x^{2}+x+1}{x+1}\to\frac{3}{2}$ . For the second one, combine over a common denominator first, because $\infty-\infty$ is never handled term by term: $\frac{1}{x}-\frac{1}{x(x+1)}=\frac{(x+1)-1}{x(x+1)}=\frac{x}{x(x+1)}=\frac{1}{x+1}\to 1$ . Two infinities that eat each other up leave a perfectly finite remainder.

d)

For $x>3$ : $|x-3|=x-3$ , the quotient equals $+1$ , so $\lim_{x\to 3^{+}}=1$ . For $x<3$ : $|x-3|=-(x-3)$ , the quotient equals $-1$ , so $\lim_{x\to 3^{-}}=-1$ . The two one-sided limits differ: the limit DOES NOT EXIST. General lesson: faced with an absolute value (or any piecewise-defined function), never compute a two-sided limit in one go: SEPARATE the two sides, remove the absolute value according to the sign, then compare. The same reflex will be used to detect points where a function is not differentiable in the next chapter.

Exercise 3: Limits at infinity and asymptotes

The figure shows the graph of $f(x)=\frac{3x^{2}-2x+1}{x^{2}-4}$ , without its asymptotes.

a) Compute $\lim_{x\to+\infty}f(x)$ and $\lim_{x\to-\infty}f(x)$ for the function in the figure, and identify the horizontal asymptote. What general method do you use?
b) Find the vertical asymptotes of $f$ and determine the sign of the limit on either side of $x=2$ .
c) Show that $g(x)=\frac{x^{2}+1}{x-1}$ has an OBLIQUE asymptote and give its equation.
d) Study $h(x)=\frac{2x}{\sqrt{x^{2}+1}}$ at $+\infty$ AND at $-\infty$ . How many horizontal asymptotes are there, and what trap does this calculation set?

Show the solution

Answers

a) $y=3$ on both sides
b) $x=\pm 2$ ; $-\infty$ to the left of $2$ , $+\infty$ to the right
c) $y=x+1$
d) $y=2$ at $+\infty$ , $y=-2$ at $-\infty$

a)

General method: divide top and bottom by the highest power present, here $x^{2}$ : $f(x)=\frac{3-\frac{2}{x}+\frac{1}{x^{2}}}{1-\frac{4}{x^{2}}}$ . As $x\to\pm\infty$ , all the terms in $\frac{1}{x}$ and $\frac{1}{x^{2}}$ vanish: the limit is $\frac{3}{1}=3$ on BOTH sides. Horizontal asymptote: $y=3$ , the horizontal dashed line in the solution figure. A legitimate shortcut for a rational function with EQUAL degrees: the ratio of the leading coefficients, here $\frac{3}{1}$ .

b)

The denominator $x^{2}-4=(x-2)(x+2)$ vanishes at $x=2$ and $x=-2$ ; there the numerator equals $12-4+1=9$ and $12+4+1=17$ respectively, both nonzero, so these are indeed two VERTICAL asymptotes (and not holes). Sign around 2: the numerator stays close to $+9$ ; the denominator $(x-2)(x+2)$ is positive for $x>2$ (a product of two positives) and negative for $x$ just below 2 (negative times positive). So $\lim_{x\to 2^{+}}f(x)=+\infty$ and $\lim_{x\to 2^{-}}f(x)=-\infty$ : the curve dives down and then comes back up on either side of the line, exactly as in the figure.

c)

The numerator has exactly ONE degree more than the denominator: that is the signature of an oblique asymptote. Divide: $x^{2}+1=(x-1)(x+1)+2$ , so $g(x)=x+1+\frac{2}{x-1}$ . As $x\to\pm\infty$ , the remainder $\frac{2}{x-1}$ tends to 0: the curve hugs the line $y=x+1$ , its oblique asymptote. The remainder even tells you the SIDE: positive for $x>1$ (the curve lies above), negative for $x<1$ (below).

d)

At $+\infty$ : $\sqrt{x^{2}+1}\approx\sqrt{x^{2}}=|x|=x$ (since $x>0$ ), so $h(x)\approx\frac{2x}{x}=2$ : asymptote $y=2$ . At $-\infty$ : careful, $\sqrt{x^{2}}=|x|=-x$ since $x<0$ ! Then $h(x)\approx\frac{2x}{-x}=-2$ : asymptote $y=-2$ . There are therefore TWO distinct horizontal asymptotes. The trap is the automatic $\sqrt{x^{2}}=x$ , false for negative $x$ : the square root always returns a POSITIVE result, hence the compulsory absolute value. Rigorously: $h(x)=\frac{2x}{|x|\sqrt{1+\frac{1}{x^{2}}}}$ , and the quotient $\frac{x}{|x|}$ equals $+1$ on the right and $-1$ on the left: it, and it alone, creates the two asymptotes.

Exercise 4: The limit laws

Limits can be computed term by term as long as no indeterminate form appears. Still, you need to know what the laws allow, and above all what they do not.

a) State the limit laws for a sum, a product, a quotient and a power, giving the hypotheses of each.
b) Compute $\lim_{x\to 2}\frac{3x^{2}-5}{x+1}$ , justifying each step by a law.
c) Give two functions $f$ and $g$ such that NEITHER $\lim_{x\to 0}f(x)$ NOR $\lim_{x\to 0}g(x)$ exists, while $\lim_{x\to 0}\left(f(x)+g(x)\right)$ does exist. Do the same for a product.
d) Why does the quotient law require the limit of the denominator to be nonzero? What can happen when it is zero?

Show the solution

Answers

a) Sum, product, power; quotient if $M\neq 0$
b) $\frac{7}{3}$
c) $s$ and $1-s$ ; $s$ and $2s$ : the converse is false
d) $\frac{k}{0}$ : infinite; $\frac{0}{0}$ : indeterminate

a)

If $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$ exist and are FINITE, then: $\lim(f+g)=L+M$ ; $\lim(fg)=LM$ ; $\lim\frac{f}{g}=\frac{L}{M}$ PROVIDED THAT $M\neq 0$ ; and $\lim f^{n}=L^{n}$ for a positive integer $n$ . It follows that $\lim(cf)=cL$ for any constant $c$ . The shared hypothesis, often left unsaid, is that BOTH limits exist and are finite: it is the one that fails in every counterexample of part c), and it is also the one missing in the indeterminate forms.

b)

The denominator first, because it decides whether the law can be used: $\lim_{x\to 2}(x+1)=3\neq 0$ , so the quotient law applies. Numerator: $\lim_{x\to 2}\left(3x^{2}-5\right)=3\lim_{x\to 2}x^{2}-5=3(4)-5=7$ , using in turn the sum, the constant multiple and the power laws. Conclusion: $\lim_{x\to 2}\frac{3x^{2}-5}{x+1}=\frac{7}{3}\approx 2.333$ . Notice that the result is simply $f(2)$ : that is the general case for a rational function at a point where the denominator does not vanish, and it is exactly what continuity means, as established in exercise 6.

c)

The simplest start is the sign function $s(x)=\frac{x}{|x|}$ , which equals $-1$ for $x<0$ and $+1$ for $x>0$ : its left-hand and right-hand limits at 0 are $-1$ and $+1$ , they differ, so $\lim_{x\to 0}s(x)$ DOES NOT EXIST. For the SUM: let $f(x)=s(x)$ and $g(x)=-s(x)+1$ . Neither has a limit at 0 (the second jumps from 2 to 0), yet $f(x)+g(x)=1$ for all $x\neq 0$ , so $\lim_{x\to 0}\left(f+g\right)=1$ exists. For the PRODUCT: let $f(x)=s(x)$ and $g(x)=2s(x)$ , still with no limit at 0; their product equals $2\left(s(x)\right)^{2}=2$ for all $x\neq 0$ , since $(\pm 1)^{2}=1$ , so $\lim_{x\to 0}(fg)=2$ exists. In both cases the lesson is the same: the converse of the laws is FALSE. The existence of the limit of a sum or a product in no way implies that the limits of the individual terms exist, because two discontinuities can cancel each other exactly. So you can never “split” an existing limit into pieces that you assume also exist: the laws read only from left to right.

d)

Because the conclusion $\frac{L}{M}$ makes no sense if $M=0$ : division by zero is undefined. When the denominator tends to 0, two situations arise. If the numerator tends to a NONZERO constant, the quotient blows up in absolute value and the sign on each side remains to be found: this is the mechanism of the vertical asymptotes in exercise 3. If the numerator ALSO tends to 0, you get the indeterminate form $\frac{0}{0}$ , where the law says nothing at all and the form must be resolved by factoring, the conjugate or a common denominator, as in exercise 2. The real content of the condition $M\neq 0$ is therefore this: it separates the cases where you can conclude mechanically from those that require algebraic work.

Exercise 5: Limits of composite functions and substitution

Setting $u=g(x)$ to reduce a complicated limit to a known one is the move that pays off most in the chapter. It rests on a theorem whose hypothesis is regularly forgotten.

a) Compute $\lim_{x\to 0}\frac{\sin 3x}{3x}$ by the substitution $u=3x$ , justifying the step.
b) Compute $\lim_{x\to 2}\sqrt{x^{2}+5}$ and $\lim_{x\to 0}e^{\frac{\sin x}{x}}$ . Which property of continuous functions are you using?
c) Compute $\lim_{x\to+\infty}x\sin\frac{1}{x}$ with a well-chosen substitution.
d) Let $f(u)=0$ if $u\neq 0$ and $f(0)=1$ , and $g(x)=0$ for all $x$ . Check that $\lim_{x\to 0}g(x)=0$ and $\lim_{u\to 0}f(u)=0$ , but that $\lim_{x\to 0}f\left(g(x)\right)\neq 0$ . Which hypothesis was missing?

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Answers

a) $1$
b) $3$ and $e$ : continuity of the outer function
c) $1$ with $u=\frac{1}{x}$
d) The composite has limit $1$ : $f$ is not continuous at $0$

a)

Let $u=3x$ . As $x\to 0$ , $u\to 0$ as well, and the expression becomes $\frac{\sin u}{u}$ . The fundamental limit gives $\lim_{u\to 0}\frac{\sin u}{u}=1$ , so $\lim_{x\to 0}\frac{\sin 3x}{3x}=1$ . The step is valid because $u=3x$ is continuous, tends to 0, and takes the value 0 only when $x$ itself does: no stray value is reached on the punctured neighbourhood. Note that the denominator is indeed $3x$ and not $x$ : had the question asked for $\lim_{x\to 0}\frac{\sin 3x}{x}$ , you would have had to write $\frac{\sin 3x}{x}=3\cdot\frac{\sin 3x}{3x}$ to recover the fundamental form, and the limit would be 3.

b)

$\lim_{x\to 2}\sqrt{x^{2}+5}=\sqrt{\lim_{x\to 2}\left(x^{2}+5\right)}=\sqrt{9}=3$ . Likewise, $\lim_{x\to 0}\frac{\sin x}{x}=1$ , so $\lim_{x\to 0}e^{\frac{\sin x}{x}}=e^{1}=e\approx 2.718$ . The property used is this one: if $g$ has limit $L$ at $a$ and $f$ is CONTINUOUS at $L$ , then $\lim_{x\to a}f\left(g(x)\right)=f(L)$ ; in other words, the limit moves inside the function. It holds because the square root and the exponential are continuous wherever they are defined; part d) shows what happens when that continuity is missing.

c)

Let $u=\frac{1}{x}$ , so that $x\to+\infty$ gives $u\to 0^{+}$ , with $x=\frac{1}{u}$ . The expression becomes $\frac{1}{u}\sin u=\frac{\sin u}{u}$ , whose limit is 1. So $\lim_{x\to+\infty}x\sin\frac{1}{x}=1$ . The substitution turned a limit at infinity, where there is no formula available for the sine, into the fundamental limit at 0: that is the typical payoff of the substitution $u=\frac{1}{x}$ , which swaps a neighbourhood of infinity for a neighbourhood of zero. Interpretation: for large $x$ , $\sin\frac{1}{x}\approx\frac{1}{x}$ , so the product is approximately 1.

d)

Checks. $g$ is the zero function, so $\lim_{x\to 0}g(x)=0$ ✓. For $f$ : when $u\to 0$ with $u\neq 0$ , we always have $f(u)=0$ , so $\lim_{u\to 0}f(u)=0$ ✓ (the limit ignores the value $f(0)=1$ , since the definition excludes the point itself). But the composite equals $f\left(g(x)\right)=f(0)=1$ for EVERY $x$ , so $\lim_{x\to 0}f\left(g(x)\right)=1\neq 0$ . The missing hypothesis is the CONTINUITY of $f$ at the limit point: here $\lim_{u\to 0}f(u)=0$ while $f(0)=1$ , so $f$ is discontinuous at 0 and the theorem of part b) does not apply. A second hypothesis would also suffice: that $g$ does NOT take the limit value 0 on a punctured neighbourhood of $a$ , which was true in part a) with $u=3x$ and false here, since $g$ is identically 0. Remember, then, that substitution is safe in two cases: a continuous outer function, or an inner function that never reaches its limit. In practice at CEGEP, all the usual functions being continuous on their domains, the first condition is almost always met, but the exercise shows that it is not a logical given.

Exercise 6: Joining the pieces: forcing continuity

A piecewise-defined function is continuous wherever each piece is; everything is decided at the JUNCTION points, where the branches are required to meet.

a) Recall the three conditions for continuity at a point, then find $a$ and $b$ so that the following function is continuous everywhere: $f(x)=x^{2}$ if $x<1$ ; $f(x)=ax+b$ if $1\le x<3$ ; $f(x)=2x$ if $x\ge 3$ .
b) Classify the discontinuity of each of these functions at $x=2$ : $u(x)=\frac{x^{2}-4}{x-2}$ ; $v(x)=\frac{1}{x-2}$ ; $w(x)=\frac{|x-2|}{x-2}$ .
c) Is the function $p(x)=\frac{x^{2}-4}{x-2}$ continuous at $x=2$ ? Can it be extended by continuity, and if so how?
d) Explain why every polynomial is continuous on $\mathbb{R}$ , and why a rational function is continuous everywhere EXCEPT at the zeros of its denominator. What practical consequence does this have for computing limits?

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a) $a=\frac{5}{2}$ , $b=-\frac{3}{2}$
b) $u$ : hole; $v$ : infinite; $w$ : jump
c) Not continuous; extended by $4$
d) Substitution wherever the denominator is nonzero

a)

Three conditions: $f(a)$ exists, $\lim_{x\to a}f(x)$ exists, and $\lim_{x\to a}f(x)=f(a)$ . Each piece is a polynomial, hence continuous on its interval: only the junctions at $x=1$ and $x=3$ are in question. At $x=1$ : the left branch tends to $1^{2}=1$ , the right branch equals $a+b$ ; equality requires $a+b=1$ . At $x=3$ : the middle branch tends to $3a+b$ , the right branch equals $6$ ; hence $3a+b=6$ . Subtracting the two equations: $2a=5$ , so $a=\frac{5}{2}$ and $b=1-\frac{5}{2}=-\frac{3}{2}$ . Check: $3\left(\frac{5}{2}\right)-\frac{3}{2}=\frac{15-3}{2}=6$ ✓.

b)

$u(x)=\frac{(x-2)(x+2)}{x-2}=x+2$ for $x\ne 2$ : the limit equals 4, finite, but $u(2)$ does not exist: a HOLE (removable discontinuity). $v(x)=\frac{1}{x-2}$ : the one-sided limits are $-\infty$ and $+\infty$ : an INFINITE discontinuity (vertical asymptote), which cannot be removed. $w(x)=\frac{|x-2|}{x-2}$ equals $-1$ on the left and $+1$ on the right: a JUMP discontinuity, which cannot be removed either. Three expressions built on the same $x-2$ , three different pathologies: it is the NUMERATOR that decides which.

c)

No: $p(2)$ does not even exist (division by zero), so the first condition fails. But the limit does exist and equals 4: so $p$ can be EXTENDED by continuity by setting $\tilde{p}(2)=4$ , which amounts to defining $\tilde{p}(x)=x+2$ everywhere. Extending by continuity means plugging a hole with the only value that reconnects the curve: the limit.

d)

A polynomial is built from the identity function and constants by sums and products; these operations PRESERVE continuity, and $\lim_{x\to a}x=a$ trivially, so by induction on the degree every polynomial is continuous everywhere. A rational function is a quotient of two continuous polynomials: the quotient stays continuous as long as the denominator does not vanish. Major practical consequence: for all these functions, $\lim_{x\to a}f(x)=f(a)$ , that is, the limit is computed by simple SUBSTITUTION. That is precisely why the only interesting exercises in the chapter are about the points where substitution fails, giving $\frac{0}{0}$ : everywhere else, there is nothing to compute.

Exercise 7: The Intermediate Value Theorem

The figure shows $f(x)=x^{3}-x-1$ on $[1,2]$ : the curve starts below the axis and ends above it, so it must cross it. The Intermediate Value Theorem turns this visual evidence into a proof.

a) State the Intermediate Value Theorem precisely, with all its hypotheses.
b) Prove that the equation $x^{3}-x-1=0$ has at least one solution in $[1,2]$ .
c) Trap this root in an interval of width $0.125$ by the BISECTION method, showing each step.
d) The function $g(x)=\frac{1}{x}$ satisfies $g(-1)=-1$ and $g(1)=1$ , yet it never takes the value 0. Is the theorem contradicted? Explain, then show that $\cos x=x$ has a solution.
e) Show that the equation $\ln x=x-4$ has at least one solution in $[0, 8]$ . Why can the theorem not be applied on $[0, 8]$ as it stands? Then show that it has a second solution, smaller than 1.

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a) $f$ continuous on $[a,b]$ , $k$ between $f(a)$ and $f(b)$
b) $f(1)=-1<0<5=f(2)$
c) $[1.25, 1.375]$
d) $g$ not continuous at $0$ ; $h(0)>0>h\left(\frac{\pi}{2}\right)$
e) $h(x)=\ln x-x+4$ continuous on $[1, 8]$ , $h(1)=3>0>h(8)$ ; not on $[0, 8]$ , since $\ln 0$ does not exist; second root in $(0.01, 1)$ .

a)

If $f$ is CONTINUOUS on the CLOSED interval $[a,b]$ and $k$ is a value between $f(a)$ and $f(b)$ , then there exists at least one $c\in[a,b]$ such that $f(c)=k$ . Two crucial points: continuity on the whole interval is essential, and the theorem asserts the EXISTENCE of such a $c$ , without saying how many there are or how to find them.

b)

$f(x)=x^{3}-x-1$ is a polynomial, hence continuous on $[1,2]$ : the hypothesis holds. Values at the endpoints: $f(1)=1-1-1=-1<0$ and $f(2)=8-2-1=5>0$ , the two points marked in the figure. The value $k=0$ does lie between $-1$ and $5$ : the theorem guarantees the existence of $c\in[1,2]$ with $f(c)=0$ . The equation therefore has at least one root in this interval, and that WITHOUT computing it.

c)

Bisection: cut the interval in half and keep the half where the sign changes. $f(1.5)=3.375-1.5-1=0.875>0$ : the root lies in $[1, 1.5]$ . $f(1.25)=1.953125-1.25-1=-0.296875<0$ : the root lies in $[1.25, 1.5]$ . $f(1.375)=2.599609-1.375-1=0.224609>0$ : the root lies in $[1.25, 1.375]$ , an interval of width $0.125$ ✓. Three steps divided the uncertainty by 8: each bisection gains a factor of 2, about three steps per decimal digit. The true root is $1.3247\ldots$ , where the curve crosses the axis in the figure.

d)

No contradiction: $g$ is NOT continuous on $[-1,1]$ , since it is not even defined at 0, where it has a vertical asymptote. The central hypothesis of the theorem collapses, so the conclusion no longer has to hold: this is precisely the counterexample showing that continuity is not a mere formality. For $\cos x=x$ : let $h(x)=\cos x-x$ , continuous everywhere (a difference of continuous functions). $h(0)=1-0=1>0$ and $h\left(\frac{\pi}{2}\right)=0-\frac{\pi}{2}\approx-1.571<0$ . By the Intermediate Value Theorem, there exists $c\in\left[0,\frac{\pi}{2}\right]$ such that $h(c)=0$ , that is, $\cos c=c$ : the equation has a solution (the famous fixed point $0.739\ldots$ ), proved in three lines although no formula gives it.

e)

Turn the equation into a root problem: $\ln x=x-4$ is equivalent to $h(x)=0$ with $h(x)=\ln x-x+4$ . The theorem cannot be applied on $[0, 8]$ as it stands: $\ln 0$ does not exist, so $h$ is neither defined nor continuous at 0, and the first hypothesis fails. Writing “ $h$ is continuous on $[0, 8]$ ” is false and costs the justification mark, even if the conclusion is right. So choose a closed interval CONTAINED in the domain $(0, +\infty)$ , for example $[1, 8]$ : $h$ is continuous there, as the sum of a logarithm and a polynomial, both continuous on $(0, +\infty)$ . We have $h(1)=0-1+4=3>0$ and $h(8)=\ln 8-4\approx 2.08-4=-1.92<0$ . Since $0$ lies between $h(8)$ and $h(1)$ , there exists $c\in\,(1, 8)$ such that $h(c)=0$ , and this $c$ is indeed a solution in $[0, 8]$ . Second solution: near 0, $\ln x\to-\infty$ , so $h$ becomes negative again; concretely $h(0.01)\approx -4.61-0.01+4=-0.62<0$ and $h(1)=3>0$ , and the theorem on $[0.01, 1]$ gives a second root in $(0.01, 1)$ . The two roots are approximately $0.019$ and $5.75$ . The choice of endpoints is the part that earns the marks: one endpoint where the value is exact, like $h(1)$ , and one where the sign is certain without a calculator, like $\ln 8<3$ since $8<e^{3}\approx 20$ .

Exercise 8: Trigonometric limits and the Squeeze Theorem

The figure shows $y=\frac{\sin x}{x}$ around 0: the function is not defined at 0 (open circle), but it tends to 1 there, the limit on which all the trigonometry of differential calculus is built.

a) State the fundamental limit illustrated by the figure and specify the angle unit it requires. What is $f(0)$ ?
b) Compute $\lim_{x\to 0}\frac{\sin 5x}{3x}$ and $\lim_{x\to 0}\frac{\tan x}{x}$ .
c) Compute $\lim_{x\to 0}\frac{1-\cos x}{x}$ , then $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}$ (multiply by the conjugate). Why are the two results so different?
d) State the Squeeze Theorem and use it to compute $\lim_{x\to 0}x^{2}\sin\frac{1}{x}$ , a limit that none of the previous techniques can reach.

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a) $\lim_{x\to 0}\frac{\sin x}{x}=1$ in radians; $f(0)$ does not exist
b) $\frac{5}{3}$ and $1$
c) $0$ and $\frac{1}{2}$
d) $0$ by the Squeeze Theorem

a)

$\lim_{x\to 0}\frac{\sin x}{x}=1$ , on the STRICT condition that $x$ is in RADIANS (in degrees, the limit would be $\frac{\pi}{180}$ : that is the real reason why all of differential calculus works in radians). As for $f(0)$ : it does not exist, since the quotient gives $\frac{0}{0}$ , hence the open circle in the figure. One more limit that exists where the function is absent.

b)

Force the pattern $\frac{\sin u}{u}$ to appear: $\frac{\sin 5x}{3x}=\frac{5}{3}\cdot\frac{\sin 5x}{5x}\to\frac{5}{3}\cdot 1=\frac{5}{3}$ (with $u=5x\to 0$ ). For the tangent: $\frac{\tan x}{x}=\frac{\sin x}{x}\cdot\frac{1}{\cos x}\to 1\cdot\frac{1}{1}=1$ . Reflex: the argument of the sine must be IDENTICAL to the denominator; adjust with a constant, never by changing the argument.

c)

Conjugate $1+\cos x$ : $\frac{1-\cos x}{x}\cdot\frac{1+\cos x}{1+\cos x}=\frac{1-\cos^{2}x}{x(1+\cos x)}=\frac{\sin^{2}x}{x(1+\cos x)}=\frac{\sin x}{x}\cdot\frac{\sin x}{1+\cos x}\to 1\cdot\frac{0}{2}=0$ . With $x^{2}$ in the denominator, the same calculation gives $\frac{\sin x}{x}\cdot\frac{\sin x}{x}\cdot\frac{1}{1+\cos x}\to 1\cdot 1\cdot\frac{1}{2}=\frac{1}{2}$ . The difference is one of SPEED: $1-\cos x$ vanishes like $x^{2}$ , not like $x$ . Divided by $x$ , a factor $x$ remains and kills everything; divided by $x^{2}$ , the balance is exact and leaves $\frac{1}{2}$ . This hierarchy ( $\sin x\sim x$ but $1-\cos x\sim\frac{x^{2}}{2}$ ) will come back intact with Taylor series.

d)

Squeeze Theorem: if $g(x)\le f(x)\le h(x)$ near $a$ and $g$ and $h$ tend to the SAME limit $L$ at $a$ , then $f$ also tends to $L$ . Here, $\sin\frac{1}{x}$ oscillates wildly and has no limit at 0, so no product law applies. But it stays bounded: $-1\le\sin\frac{1}{x}\le 1$ , hence, multiplying by $x^{2}\ge 0$ : $-x^{2}\le x^{2}\sin\frac{1}{x}\le x^{2}$ . Both bounds tend to 0: the function is crushed between them, so $\lim_{x\to 0}x^{2}\sin\frac{1}{x}=0$ . The Squeeze Theorem is the tool for oscillating functions: you do not try to follow the oscillations, you TRAP them.

Exercise 9: The formal definition: epsilon and delta

This is where the CEGEP course goes further than high school: the limit stops being an intuition of getting closer and becomes a quantified, checkable statement.

a) State the formal definition of $\lim_{x\to a}f(x)=L$ with the quantifiers, then translate it into plain English.
b) Prove that $\lim_{x\to 3}(2x-1)=5$ by exhibiting an explicit $\delta$ in terms of $\varepsilon$ .
c) Prove that $\lim_{x\to 2}x^{2}=4$ . Why do you need to BOUND $\delta$ at the start here, unlike in part b)?
d) Why does the definition require $0<|x-a|$ and not simply $|x-a|<\delta$ ? Which example from the chapter would be destroyed by leaving it out?

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a) $\forall\varepsilon>0,\exists\delta>0$ : $0<|x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$
b) $\delta=\frac{\varepsilon}{2}$
c) $\delta=\min\left(1,\frac{\varepsilon}{5}\right)$
d) The limit ignores the value at $a$

a)

Definition: $\lim_{x\to a}f(x)=L$ means that for every $\varepsilon>0$ , there exists $\delta>0$ such that, for all $x$ , $0<|x-a|<\delta\;\Rightarrow\;|f(x)-L|<\varepsilon$ . In plain English: however narrow the horizontal band of half-width $\varepsilon$ drawn around $L$ , you can always find a vertical band of half-width $\delta$ around $a$ such that the curve, over that vertical band (the point $a$ excepted), stays entirely inside the horizontal band. The order of the quantifiers is essential: the opponent chooses $\varepsilon$ FIRST, and you must supply a $\delta$ that depends on it.

b)

Let $\varepsilon>0$ be given. Start from the quantity to control: $|f(x)-L|=|(2x-1)-5|=|2x-6|=2|x-3|$ . We want $2|x-3|<\varepsilon$ , which is equivalent to $|x-3|<\frac{\varepsilon}{2}$ . So it suffices to take $\delta=\frac{\varepsilon}{2}$ . Check: if $0<|x-3|<\delta=\frac{\varepsilon}{2}$ , then $|f(x)-5|=2|x-3|<2\cdot\frac{\varepsilon}{2}=\varepsilon$ ✓. The proof is complete for every $\varepsilon>0$ . A concrete example: for $\varepsilon=0.01$ , take $\delta=0.005$ . The general method for a linear function is always this one: factor out $|x-a|$ and read off the coefficient.

c)

Let $\varepsilon>0$ . Factor: $\left|x^{2}-4\right|=|x-2|\cdot|x+2|$ . The factor $|x-2|$ is the one we control, but $|x+2|$ still depends on $x$ : it must be BOUNDED by a constant. To do so, first require $\delta\leq 1$ , which guarantees $|x-2|<1$ , so $1<x<3$ , so $3<x+2<5$ and $|x+2|<5$ . Then $\left|x^{2}-4\right|<5|x-2|$ , and it suffices to have $5|x-2|<\varepsilon$ , that is, $|x-2|<\frac{\varepsilon}{5}$ . So take $\delta=\min\left(1,\frac{\varepsilon}{5}\right)$ ✓. You must bound $\delta$ at the start, unlike in part b), because the function is no longer linear: the “coefficient” $|x+2|$ varies with $x$ , and without a prior constraint it could be arbitrarily large, making any uniform bound impossible. Example: for $\varepsilon=0.1$ , take $\delta=\min(1, 0.02)=0.02$ , and check that $|x-2|<0.02$ implies $\left|x^{2}-4\right|<0.02\times 4.02=0.0804<0.1$ ✓.

d)

The condition $0<|x-a|$ explicitly excludes the point $x=a$ itself: it says that the limit depends ONLY on the behaviour of $f$ near $a$ , never on its value AT $a$ . Without this exclusion, the definition would require $|f(a)-L|<\varepsilon$ for every $\varepsilon$ , hence $f(a)=L$ , and “having a limit” would become a synonym of “being continuous”. The whole chapter would collapse. The first example destroyed would be $\lim_{x\to 2}\frac{x^{2}-4}{x-2}=4$ from exercise 2: the function is not even DEFINED at 2, and yet its limit exists. With the wrong definition, no function with a hole would have a limit, the derivative itself could not be defined (the difference quotient $\frac{f(x)-f(a)}{x-a}$ is never defined at $x=a$ ), and there would be no differential calculus. That little $0<$ holds up the whole building.

Exercise 10: Continuity on an interval and the Extreme Value Theorem

Continuity at a single point gives nothing on its own. It is continuity on a CLOSED and BOUNDED interval that produces the great theorems, and that will justify the whole optimization method.

a) What does “ $f$ is continuous on $[a,b]$ ” mean? Why do the two endpoints receive special treatment?
b) State the Extreme Value Theorem with all its hypotheses.
c) Show with three examples that the conclusion fails if one of the hypotheses is removed: an unbounded interval, an interval that is not closed, a discontinuous function.
d) Explain why this theorem justifies the optimization method on a closed interval, and what must be done instead when the domain is open.

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a) Continuous from the right at $a$ , from the left at $b$
b) Continuous on $[a,b]$ , closed and bounded: the bounds are attained
c) $x$ on $[0,+\infty)$ , $x$ on $(0,1)$ , $\frac{1}{x}$ extended by $0$
d) Existence guaranteed; on an open interval, sign of $f'$

a)

$f$ is continuous on $[a,b]$ if it is continuous at every INTERIOR point $c\in\left(a,b\right)$ in the usual sense ( $\lim_{x\to c}f(x)=f(c)$ ), and if it is continuous FROM THE RIGHT at $a$ ( $\lim_{x\to a^{+}}f(x)=f(a)$ ) and FROM THE LEFT at $b$ ( $\lim_{x\to b^{-}}f(x)=f(b)$ ). The endpoints receive special treatment because the two-sided limit makes no sense there: the function is not defined on the other side. Requiring $\lim_{x\to a}f(x)=f(a)$ in the full sense would be a condition impossible to meet, so we settle for the one-sided limit on the side where the function exists.

b)

EXTREME VALUE THEOREM: if $f$ is continuous on a CLOSED and BOUNDED interval $[a,b]$ , then $f$ is bounded there and it ATTAINS its bounds, that is, there exist $c$ and $d$ in $[a,b]$ such that $f(c)\leq f(x)\leq f(d)$ for all $x\in[a,b]$ . Three hypotheses are therefore at stake: continuity, the interval being closed (the endpoints belong to it) and the interval being bounded. The theorem is an EXISTENCE theorem: it asserts that the maximum and the minimum exist, without saying where they are.

c)

UNBOUNDED interval: $f(x)=x$ on $\left[0,+\infty\right)$ is continuous, but has no maximum, since it grows without bound. Interval NOT CLOSED: $f(x)=x$ on $\left(0,1\right)$ is continuous and bounded, but attains neither a maximum nor a minimum, since the values 0 and 1 it approaches are never taken; worse, $f(x)=\frac{1}{x}$ on $\left(0,1\right]$ is continuous and not even bounded. DISCONTINUOUS function: on $[0,1]$ , let $f(x)=\frac{1}{x}$ for $x\neq 0$ and $f(0)=0$ ; the interval is indeed closed and bounded, but the function is not continuous at 0 and it is not bounded. The three hypotheses are therefore INDEPENDENT and all necessary: none of them is decorative.

d)

The optimization method on $[a,b]$ consists of evaluating $f$ at the critical numbers and at the two endpoints, then keeping the largest and the smallest value. This method is valid only because the theorem GUARANTEES that the maximum and the minimum exist: without it, you would be comparing a list of candidates without knowing whether the winner is really an absolute extremum, or even whether one exists. The theorem provides existence, and the analysis with the derivative provides location, since an extremum on a closed interval is attained either at a critical number or at an endpoint. When the domain is OPEN or unbounded, the theorem no longer applies and you must justify differently: by the sign of $f'$ on the whole interval (increasing then decreasing around a single critical number), by concavity through $f''$ , or by showing that $f$ tends to infinity (or to a worse value) at the ends of the domain. That is exactly the situation of the open-top box in the applied optimization set, whose domain $\left(0,+\infty\right)$ rules out any use of the endpoints.

Exercise 11: Finding the parameters that make a function continuous

Consider the following functions, where $a$ , $b$ , $c$ , $d$ and $m$ are real constants.

$f(x)=\frac{\sqrt{x+4}-2}{x}$ if $x>0$ ; $f(0)=a$ ; $f(x)=\frac{\sin(bx)}{x}$ if $x<0$ , with $b\neq 0$ .

$g(x)=\frac{x^{2}+cx-6}{x-2}$ if $x\neq 2$ ; $g(2)=d$ .

a) Find $a$ and $b$ so that $f$ is continuous at $0$ .
b) Find $c$ so that $\lim_{x\to 2}g(x)$ is finite, then $d$ so that $g$ is continuous at $2$ .
c) Find $m$ so that $h(x)=x^{2}-m$ if $x\le 1$ , $h(x)=mx+2$ if $x>1$ , is continuous on $\mathbb{R}$ .
d) Can a value $p(0)$ be chosen that makes the function $p(x)=\frac{1}{x}$ continuous at $0$ ? Justify.

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Answers

a) $a=b=\frac{1}{4}$
b) $c=1$ , $d=5$
c) $m=-\frac{1}{2}$
d) No: infinite discontinuity

a)

Continuity at $0$ requires the two one-sided limits to exist, to be equal, and to equal $f(0)=a$ . On the right, a form $\frac{0}{0}$ resolved by the conjugate: $\frac{\sqrt{x+4}-2}{x}=\frac{(x+4)-4}{x\left(\sqrt{x+4}+2\right)}=\frac{1}{\sqrt{x+4}+2}\to\frac{1}{4}$ . On the left, bring out the pattern $\frac{\sin u}{u}$ : $\frac{\sin(bx)}{x}=b\cdot\frac{\sin(bx)}{bx}\to b$ . So $b=\frac{1}{4}$ and $a=\frac{1}{4}$ . The three conditions bear on three different objects, two limits and one value, and each gives an equation.

b)

The denominator tends to $0$ . If the numerator tended to a nonzero number, the limit would be infinite: for a FINITE limit, the numerator must therefore vanish at $2$ , that is, $4+2c-6=0$ and $c=1$ . Then $x^{2}+x-6=(x-2)(x+3)$ and $g(x)=x+3$ for $x\neq 2$ , with limit $5$ at $2$ . For continuity, $d=5$ . This argument, “the numerator must vanish where the denominator vanishes”, is a NECESSARY condition: you must then check, by factoring, that the limit is indeed finite.

c)

Each piece is a polynomial, continuous on its interval: only the junction at $1$ matters. $h(1)=1-m$ and $\lim_{x\to 1^{-}}h(x)=1-m$ ; $\lim_{x\to 1^{+}}h(x)=m+2$ . Setting $1-m=m+2$ gives $m=-\frac{1}{2}$ . Check: $1+\frac{1}{2}=\frac{3}{2}$ and $-\frac{1}{2}+2=\frac{3}{2}$ ✓.

d)

No. $\lim_{x\to 0^{+}}\frac{1}{x}=+\infty$ and $\lim_{x\to 0^{-}}\frac{1}{x}=-\infty$ : no finite limit exists at $0$ , and no value $p(0)$ can equal a limit that does not exist. The discontinuity is INFINITE, hence not removable. Only a hole, where the limit exists and is finite, can be removed by redefining one point.

Exercise 12: The floor function: one-sided limits, continuity and squeezing

$\lfloor x\rfloor$ denotes the floor of $x$ (its integer part), the greatest integer less than or equal to $x$ : $\lfloor 2.7\rfloor=2$ and $\lfloor -1.3\rfloor=-2$ . The figure shows its graph: solid dots are included, open circles excluded.

a) Compute $\lim_{x\to 2^{-}}\lfloor x\rfloor$ , $\lim_{x\to 2^{+}}\lfloor x\rfloor$ and $\lfloor 2\rfloor$ . Is the function continuous at $2$ ?
b) At an integer $n$ , is the floor function continuous from the left, from the right, or neither?
c) Let $u(x)=\lfloor x\rfloor+\lfloor -x\rfloor$ . Compute $u(2.5)$ , $u(2)$ and $\lim_{x\to 2}u(x)$ . Name the discontinuity.
d) Let $v(x)=(x-1)\lfloor x\rfloor$ . Show that $v$ is continuous at $1$ , while $\lfloor x\rfloor$ is not.
e) Bound $\frac{\lfloor x\rfloor}{x}$ for $x>1$ and compute $\lim_{x\to+\infty}\frac{\lfloor x\rfloor}{x}$ .

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Answers

a) $1$ and $2$ ; $\lfloor 2\rfloor=2$ : discontinuous
b) Continuous from the right only
c) $-1$ ; $0$ ; limit $-1$ : hole
d) Limits and value all zero: continuous
e) $1-\frac{1}{x}<\frac{\lfloor x\rfloor}{x}\le 1$ : limit $1$

a)

Just to the left of $2$ , for example at $1.99$ , the floor equals $1$ : $\lim_{x\to 2^{-}}\lfloor x\rfloor=1$ . Just to the right, it equals $2$ : $\lim_{x\to 2^{+}}\lfloor x\rfloor=2$ , and $\lfloor 2\rfloor=2$ . The one-sided limits differ: the limit does not exist and the function is discontinuous at $2$ , with a jump of height $1$ .

b)

At $n$ , the right-hand limit equals $n=\lfloor n\rfloor$ , the left-hand limit equals $n-1$ . The floor function is therefore continuous FROM THE RIGHT at every integer, never from the left: that is what the solid dot at the start of each step says. Between two integers it is constant, hence continuous.

c)

$u(2.5)=2+\lfloor -2.5\rfloor=2+(-3)=-1$ . $u(2)=2+(-2)=0$ . For non-integer $x$ , $\lfloor -x\rfloor=-\lfloor x\rfloor-1$ , so $u(x)=-1$ : this holds on both sides of $2$ , and $\lim_{x\to 2}u(x)=-1$ . The limit exists, the value exists, they differ: a HOLE discontinuity, removable by setting $u(2)=-1$ . Two step functions have added up to an almost constant function, whose jumps cancel exactly.

d)

$v(1)=0\times 1=0$ . On the left, $\lfloor x\rfloor=0$ and $v(x)=0$ : limit $0$ . On the right, $\lfloor x\rfloor=1$ and $v(x)=x-1\to 0$ . All three numbers are $0$ : $v$ is continuous at $1$ . The factor $x-1$ , which vanishes exactly at the point of the jump, flattens the discontinuity of the second factor. The “product of continuous functions” rule does not apply, and yet the conclusion is true: the rules give sufficient conditions, not necessary ones.

e)

By definition, $x-1<\lfloor x\rfloor\le x$ . For $x>1$ , divide by $x>0$ without reversing the inequalities: $1-\frac{1}{x}<\frac{\lfloor x\rfloor}{x}\le 1$ . Both bounds tend to $1$ as $x\to+\infty$ : by the Squeeze Theorem, $\lim_{x\to+\infty}\frac{\lfloor x\rfloor}{x}=1$ . The function itself never simplifies, it oscillates below $1$ in a sawtooth, but there is no need to follow it: you trap it.

Exercise 13: Limits of exponential, logarithmic and arctangent functions

None of these limits requires L'Hôpital's rule: each one is read off the behaviour of a basic function, provided you know from which side you are coming.

a) Compute $\lim_{x\to 0^{-}}e^{\frac{1}{x}}$ and $\lim_{x\to 0^{+}}e^{\frac{1}{x}}$ .
b) Compute $\lim_{x\to 0^{-}}\arctan\left(\frac{1}{x}\right)$ and $\lim_{x\to 0^{+}}\arctan\left(\frac{1}{x}\right)$ . What type of discontinuity do you get?
c) Compute $\lim_{x\to+\infty}\frac{3\cdot 2^{x}-5^{x}}{5^{x}+4}$ and $\lim_{x\to-\infty}\frac{3\cdot 2^{x}-5^{x}}{5^{x}+4}$ .
d) Compute $\lim_{x\to+\infty}\left[\ln(2x+1)-\ln(x+3)\right]$ .
e) Give the asymptotes of $k(x)=\frac{2e^{x}+1}{e^{x}-1}$ , with the sign of the limits on either side of the vertical asymptote.

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Answers

a) $0$ and $+\infty$
b) $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ : jump
c) $-1$ and $0$
d) $\ln 2\approx 0.693$
e) $y=2$ , $y=-1$ , $x=0$ ; $-\infty$ on the left, $+\infty$ on the right

a)

As $x\to 0^{-}$ , $\frac{1}{x}\to-\infty$ , and $e^{t}\to 0$ as $t\to-\infty$ : $\lim_{x\to 0^{-}}e^{\frac{1}{x}}=0$ . As $x\to 0^{+}$ , $\frac{1}{x}\to+\infty$ and $e^{\frac{1}{x}}\to+\infty$ . Same expression, two sides, two opposite behaviours: here a limit at $0$ without specifying the side means nothing.

b)

On the left, $\frac{1}{x}\to-\infty$ and $\arctan t\to-\frac{\pi}{2}$ : limit $-\frac{\pi}{2}$ . On the right, $+\frac{\pi}{2}$ . The two limits are FINITE and different: this is a jump discontinuity, of height $\pi$ , and no value at $0$ can remove it.

c)

At $+\infty$ , the dominant term is $5^{x}$ , in both the numerator and the denominator. Divide everything by $5^{x}$ : $\frac{3\left(\frac{2}{5}\right)^{x}-1}{1+4\cdot 5^{-x}}$ . Now $\left(\frac{2}{5}\right)^{x}\to 0$ and $5^{-x}\to 0$ : the limit is $-1$ . At $-\infty$ , it is the reverse: $2^{x}\to 0$ and $5^{x}\to 0$ , the numerator tends to $0$ and the denominator to $4$ , limit $0$ . The largest base dominates at $+\infty$ , the constant dominates at $-\infty$ : you must sort the terms again at each end.

d)

Each logarithm tends to $+\infty$ : a form $\infty-\infty$ . Combine them before taking the limit: $\ln\left(\frac{2x+1}{x+3}\right)$ . The quotient tends to $2$ , and the logarithm is continuous at $2$ : the limit is $\ln 2\approx 0.693$ .

e)

At $+\infty$ , divide by $e^{x}$ : $\frac{2+e^{-x}}{1-e^{-x}}\to 2$ , asymptote $y=2$ . At $-\infty$ , $e^{x}\to 0$ : $k(x)\to\frac{0+1}{0-1}=-1$ , asymptote $y=-1$ . The denominator vanishes at $x=0$ , where the numerator equals $3$ : vertical asymptote $x=0$ . As $x\to 0^{+}$ , $e^{x}>1$ , the denominator is positive and small: $+\infty$ . As $x\to 0^{-}$ , it is negative: $-\infty$ . Two DIFFERENT horizontal asymptotes, as for functions with a square root, but for another reason: the exponential behaves completely differently at the two ends.

Exercise 14: Problem: an operator's learning curve

An operator is starting on a new machine. Their output rate, in parts per hour, is modelled by $P(t)=40-30e^{-0.25t}$ , where $t$ is the number of days worked since the start of their training.

After $20$ days worked, they go on two weeks of leave. On their return, their output rate is modelled by $Q(t)=40-6.2e^{-0.5(t-20)}$ for $t\ge 20$ , where $t$ still counts the days worked.

a) Compute $P(0)$ and $\lim_{t\to+\infty}P(t)$ . Interpret these results.
b) Compute $P(5)$ to one decimal place.
c) Use the Intermediate Value Theorem to show that the output rate reaches $35$ parts per hour between day $5$ and day $10$ , then find that time exactly.
d) Compute $\lim_{t\to 20^{-}}P(t)$ and $Q(20)$ . Is the output rate, described by $P$ and then by $Q$ , continuous at $t=20$ ? Quantify the gap.
e) After how many days worked since the start does the operator get back to a rate of $39$ parts per hour?

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Answers

a) $10$ parts per hour; ceiling of $40$
b) $31.4$
c) $t=4\ln 6\approx 7.17$ days
d) $39.80$ against $33.8$ : a jump of $6.0$
e) $t\approx 23.65$

a)

$P(0)=40-30=10$ parts per hour on the first day. As $t\to+\infty$ , $e^{-0.25t}\to 0$ and $P(t)\to 40$ : the output rate levels off at $40$ parts per hour. The line $y=40$ is a horizontal asymptote that the model never reaches: it gets ever closer, which is a good description of learning.

b)

$P(5)=40-30e^{-1.25}\approx 40-30\times 0.2865\approx 31.4$ parts per hour.

c)

$P$ is continuous on $\left[5,10\right]$ , like any combination of exponentials. $P(5)\approx 31.4<35$ and $P(10)=40-30e^{-2.5}\approx 37.5>35$ : by the Intermediate Value Theorem, there exists $c\in\left[5,10\right]$ such that $P(c)=35$ . To find it: $30e^{-0.25t}=5$ , so $e^{-0.25t}=\frac{1}{6}$ and $t=4\ln 6\approx 7.17$ days. The theorem guaranteed existence; the logarithm gives the value.

d)

$\lim_{t\to 20^{-}}P(t)=40-30e^{-5}\approx 39.80$ , since $P$ is continuous. $Q(20)=40-6.2=33.8$ . The two differ by $6.0$ parts per hour: the output rate has a JUMP at $t=20$ ; the leave cost six parts per hour. The right-hand limit exists and equals $Q(20)$ , the left-hand limit also exists, but they do not agree.

e)

On return, $Q(t)=39$ gives $6.2e^{-0.5(t-20)}=1$ , so $e^{-0.5(t-20)}=\frac{1}{6.2}$ and $t-20=2\ln 6.2\approx 3.65$ . The operator gets back to $39$ parts per hour around $t\approx 23.65$ , that is, during the 24th day worked. Before the leave, they had reached that level when $30e^{-0.25t}=1$ , that is, $t=4\ln 30\approx 13.6$ days: relearning is much faster than learning, which is what the coefficient $0.5$ against $0.25$ expresses.

Exercise 15: Problem: the cost of pollution removal and the vertical asymptote

A plant must remove part of a pollutant from its discharge. The cost, in thousands of dollars, of removing $p$ percent of it is modelled by $C(p)=\frac{120p}{100-p}$ .

a) Give the domain of $C$ in context and compute $C(0)$ .
b) Compute $C(50)$ and $C(90)$ .
c) Compute $\lim_{p\to 100^{-}}C(p)$ . Why does only one one-sided limit make sense, and what does this result say?
d) The budget is $600$ thousand dollars. What percentage can be removed?
e) Compare the cost of going from $0$ to $90$ percent with the cost of going from $90$ to $99$ percent.

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Answers

a) $[0,100)$ ; $C(0)=0$
b) $120$ and $1080$ thousand dollars
c) $+\infty$ : removing everything is impossible
d) $83.3$ percent
e) $1080$ against $10\,800$ thousand dollars

a)

A percentage removed ranges from $0$ to $100$ , and $p=100$ makes the denominator zero: domain $\left[0,100\right)$ . $C(0)=0$ : removing nothing costs nothing.

b)

$C(50)=\frac{6000}{50}=120$ thousand dollars. $C(90)=\frac{10800}{10}=1080$ thousand dollars. Going from half to nine tenths multiplies the cost by nine.

c)

As $p\to 100^{-}$ , the numerator tends to $12000$ and the denominator to $0$ through POSITIVE values: $C(p)\to+\infty$ . The right-hand limit makes no sense, since you cannot remove more than $100$ percent, and the formula would give a negative cost there anyway. The vertical asymptote $p=100$ expresses an economic reality: removing ALL of the pollutant is impossible at a finite cost, each extra point costing more than the previous one.

d)

$\frac{120p}{100-p}=600$ gives $120p=60000-600p$ , so $720p=60000$ and $p\approx 83.3$ percent. Check that $83.3$ is in the domain. Since $C$ is continuous and increasing on $\left[0,100\right)$ , every positive budget corresponds to one and only one percentage: that is the Intermediate Value Theorem, together with monotonicity.

e)

From $0$ to $90$ : $1080$ thousand dollars. $C(99)=\frac{11880}{1}=11880$ , so from $90$ to $99$ : $11880-1080=10800$ thousand dollars, ten times as much for nine percentage points. Near the asymptote, the marginal cost explodes: the last percent, from $99$ to $100$ , has no price.

Corrected exercises: limits and continuity (201-SN2-RE)

Course recap

Calculations: limits at a point

Exercise 0.1: Computing a limit at a point: twelve direct calculations

Calculations: limits at infinity

Exercise 0.2: Computing a limit at infinity: ten direct calculations

Part A: computing limits (/50)

Exercise 1: Reading limits from a graph

Exercise 2: Resolving indeterminate forms

Exercise 3: Limits at infinity and asymptotes

Exercise 4: The limit laws

Exercise 5: Limits of composite functions and substitution

Part B: rigour, continuity and theorems (/50)

Exercise 6: Joining the pieces: forcing continuity

Exercise 7: The Intermediate Value Theorem

Exercise 8: Trigonometric limits and the Squeeze Theorem

Exercise 9: The formal definition: epsilon and delta

Exercise 10: Continuity on an interval and the Extreme Value Theorem

Part C: the classics (/50)

Exercise 11: Finding the parameters that make a function continuous

Exercise 12: The floor function: one-sided limits, continuity and squeezing

Exercise 13: Limits of exponential, logarithmic and arctangent functions

Exercise 14: Problem: an operator's learning curve

Exercise 15: Problem: the cost of pollution removal and the vertical asymptote

See also

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