Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: limits and continuity (201-SN2-RE)

Q: What is the difference between the limit of a function at a point and its value at that point?

The limit describes what the function does around the point; the value is what it equals at the point. The two can both exist and be different, and the limit can perfectly well exist where the function is not defined. This is exactly the distinction that every question on continuity tests.

Q: How do you resolve a 0 over 0 indeterminate form?

You transform the expression, you do not compute. With polynomials, factor and cancel the factor that vanishes. With a square root, multiply by the conjugate expression. With nested fractions, combine over a common denominator. An indeterminate form is not an answer; it is an order to change the form.

Q: How do you show that a function is continuous at a point?

Three conditions, in this order, and all three written on your paper: the function is defined at the point, the limit exists at that point, and these two numbers are equal. For a piecewise-defined function, compute the two one-sided limits at the junction point, then set them equal to the required value.

Q: When do you use the Intermediate Value Theorem?

As soon as the question asks you to show that an equation has a solution on an interval. Check continuity on the closed interval, compute the values at the two endpoints, note the change of sign, then conclude that a solution exists. Uniqueness never comes from this theorem: it is justified separately, by strict monotonicity.

Q: Why can a square root give two horizontal asymptotes?

Because the square root of x squared equals the absolute value of x: that gives x at plus infinity, but minus x at minus infinity. The sign changes with the side, so the two limits at infinity can be opposite. It is the classic trap for functions with a square root in the numerator or the denominator.

Q: Can a function be continuous on one interval and discontinuous on another containing the same point?

Yes, and that is exactly what the first test checks. At a closed endpoint, continuity is required on one side only, the side where the interval lies. A jump at the right endpoint of an interval therefore does not stop the function from being continuous on that interval, while the same jump in the middle of a larger interval makes it discontinuous, and an interval that is open at that point makes it continuous again. The question is never whether the function is continuous, but whether it is continuous on that particular interval.

This sheet does not repeat the course on limits: you already have it in your notes. It deals with what the course never says and yet decides the grade: the exact error that costs two marks on a test, the method to choose according to the FORM of the indeterminate expression, and the write-up that the grader of the first 201-SN2-RE test expects word for word.

It first gives the tables of limit laws, for sums, products, quotients and powers, and the comparison of degrees, each line followed by a worked numerical example, then the traps of the first test. It is written for CEGEP students in the Science program taking Differential Calculus, 201-SN2-RE (formerly 201-NYA). Once you have read it, the corrected exercise set on the same chapter puts each reflex to the test.

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The thread of the chapter

A limit describes what the function does AROUND the point, never at the point. Almost all the marks lost in this chapter come from confusing $\lim_{x\to a}f(x)$ with $f(a)$ , or from treating $\infty$ as a number.

This chapter is part of 201-SN2-RE, Differential Calculus

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Limit of a function
One-sided limits
Indeterminate forms
Limit at infinity
Vertical and horizontal asymptotes
Continuity at a point
Piecewise-defined function
Intermediate Value Theorem
Squeeze Theorem
Special trigonometric limits
Epsilon-delta definition of a limit
Continuity on an interval

The essentials

Three different questions at a single point

• $\lim_{x\to a^{-}}f(x)$ and $\lim_{x\to a^{+}}f(x)$ : where the function is heading as you come from the left, then from the right.
• $\lim_{x\to a}f(x)$ : it exists IF AND ONLY IF both one-sided limits exist and are equal.
• $f(a)$ : the value at the point. It can differ from both, or not exist at all, without changing anything about the limit.
• A test often asks for all three at the same $a$ on a graph: three answers, not one.

Three readings at the same point

x=2

: the curve rises towards

3

on the left, comes back down from

2

on the right, and the solid dot gives

f(2)=4

. Open circles are not values of

f

, they are directions.

Remember the formula that sums up the chapter: the limit looks at the neighbourhood, continuity compares the neighbourhood with the point.

The seven indeterminate forms, and the ones that are not

• Indeterminate: $\frac{0}{0}$ , $\frac{\infty}{\infty}$ , $\infty-\infty$ , $0\cdot\infty$ , $1^{\infty}$ , $\infty^{0}$ , $0^{0}$ .
• NOT indeterminate: $\frac{5}{0^{+}}=+\infty$ , $\frac{0}{5}=0$ , $\frac{5}{\infty}=0$ , $\infty+\infty=+\infty$ .
• An indeterminate form is not a result: it is the observation that the form does not decide, so the expression must be transformed.

Writing “indeterminate form $\frac{0}{0}$ ” before transforming often earns half a mark: it proves you know why you are factoring.

Continuity at $a$ : three conditions, in this order

• 1. $f(a)$ exists (the point is in the domain).
• 2. $\lim_{x\to a}f(x)$ exists (the two one-sided limits agree).
• 3. $\lim_{x\to a}f(x)=f(a)$ .
• If 2 holds but 3 fails: a HOLE, removable by redefining $f(a)$ .
• If both one-sided limits exist and differ: a JUMP, not removable.
• If a one-sided limit is infinite: an INFINITE discontinuity, with a vertical asymptote.

Hole at

x=1

: the limit exists but differs from

f(1)

, removed by redefining the point. Jump at

x=5

2.5

on the left,

4.5

on the right. Infinite at

x=8

: a vertical asymptote.

Continuity ON AN INTERVAL: the endpoint changes the answer

• In the INTERIOR of the interval: ordinary continuity, the three conditions at every point.
• At a CLOSED endpoint: only ONE-SIDED continuity, on the side where the interval lies. At $b$ for $[a,b]$ , it is the left-hand limit that must equal $f(b)$ ; the right side is not looked at.
• At an OPEN endpoint: nothing at all, the point does not belong to the interval.
• Consequence: the same function, the same break point, four answers. If $f$ jumps at $c$ while keeping $\lim_{x\to c^{-}}f(x)=f(c)$ , then $f$ is continuous on $[a,c]$ , DISCONTINUOUS on $[c,b]$ , continuous on $(c,b)$ , and discontinuous on $[a,b]$ , where $c$ is an interior point.

The same point

x=2

gives four verdicts: continuous on

[0, 2]

, where the left-hand limit does equal

f(2)=2

, discontinuous on

[2, 5]

and on

[0, 5]

, continuous again on

(2, 5]

, which excludes the point.

A test never asks “is $f$ continuous?” but “is $f$ continuous ON ...?”. Reading the brackets is part of the question: it is the only place in the chapter where the answer depends on them.

Limits to recognize at a glance

• $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}$ , in RADIANS, and only when the argument of the sine is exactly the denominator.
• Rational function at $\pm\infty$ : lower degree on top, the limit is $0$ ; equal degrees, it is the ratio of the leading coefficients; degree on top higher by exactly one, there is an oblique asymptote found by long division.
• $\sqrt{x^{2}}=|x|$ , so $x$ at $+\infty$ but $-x$ at $-\infty$ .

\frac{\sin x}{x}

is not defined at

0

, but the curve arrives at

(0, 1)

from both sides: the limit is

1

(in radians). There is only a hole, no asymptote.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Sum: the limit of $f+g$

Substitute, read the two limits, find the line. Only one line is red, and it is the one that comes back on every test.

$\lim f$	$\lim g$	$\lim (f+g)$
$L$	$L'$	$L+L'$
Example: $\lim_{x\to 4}\left(\sqrt{x}+x\right)=2+4=6$
$L$	$+\infty$	$+\infty$
Example: $\lim_{x\to 0^{+}}\left(5+\frac{1}{x}\right)=5+\infty=+\infty$
$L$	$-\infty$	$-\infty$
Example: $\lim_{x\to 0^{+}}\left(2-\frac{1}{x}\right)=2-\infty=-\infty$
$+\infty$	$+\infty$	$+\infty$
Example: $\lim_{x\to+\infty}\left(x^{3}+7x\right)=+\infty$ , with no calculation: 2 quantities blowing up upwards cannot cancel each other.
$-\infty$	$-\infty$	$-\infty$
Example: $\lim_{x\to-\infty}\left(x^{3}+x\right)=-\infty$
$+\infty$	$-\infty$	$\infty-\infty$ settles nothing
Example: $\lim_{x\to 2}\left(\frac{1}{x-2}-\frac{4}{x^{2}-4}\right)=\lim_{x\to 2}\frac{x-2}{(x-2)(x+2)}=\frac{1}{4}$ , after combining over a common denominator. Same form, other result: $\lim_{x\to+\infty}\left(x^{2}-x\right)=+\infty$ : the same form $\infty-\infty$ , and this time the result is infinite. What to do: Three moves depending on the expression: combine over a common denominator if there are fractions, multiply by the conjugate if there is a square root, factor out the dominant term between polynomials.

The two calculations on the red line have the same form and two unrelated answers, $\frac{1}{4}$ and $+\infty$ . That is the definition of an indeterminate form: the form decides nothing, it orders you to rewrite.

Product: the limit of $f\times g$

The sign rule applies as it does for numbers. The only indeterminate case is when one factor tends to $0$ while the other blows up: $0$ has no sign, so it cannot settle the matter.

$\lim f$	$\lim g$	$\lim (f\times g)$
$L$	$L'$	$L\times L'$
Example: $\lim_{x\to 4}\sqrt{x}\,(x-1)=2\times 3=6$
$L>0$	$+\infty$	$+\infty$
Example: $\lim_{x\to 0^{+}}(x+3)\cdot\frac{1}{x^{2}}=3\times(+\infty)=+\infty$
$L<0$	$+\infty$	$-\infty$
Example: $\lim_{x\to 0^{+}}(x-3)\cdot\frac{1}{x^{2}}=(-3)\times(+\infty)=-\infty$
$\pm\infty$	$\pm\infty$	$\pm\infty$ , sign rule
Example: $\lim_{x\to 0^{+}}\left(-\frac{1}{x}\right)\cdot\frac{1}{x^{2}}=(-\infty)\times(+\infty)=-\infty$
$0$	$\pm\infty$	$0\cdot\infty$ settles nothing
Example: $\lim_{x\to 2}(x-2)\cdot\frac{1}{x^{2}-4}=\lim_{x\to 2}\frac{1}{x+2}=\frac{1}{4}$ Same form, other result: $\lim_{x\to 2^{+}}(x-2)\cdot\frac{1}{(x-2)^{2}}=\lim_{x\to 2^{+}}\frac{1}{x-2}=+\infty$ : the same form $0\cdot\infty$ , an infinite result. What to do: Write the product as a SINGLE quotient, then cancel the common factor. This is always possible, and it brings the line back to the table of quotients.

On lines 2 and 3, it is the SIGN of the finite limit that decides, not the sign of the function before taking the limit. A function that is negative on a whole interval can very well have a positive limit.

Quotient: the limit of $\frac{f}{g}$

Only two lines are indeterminate. The truly dangerous lines are not: they are the ones where the denominator tends to $0$ , where everything hinges on the sign of that zero, and where vertical asymptotes are decided.

$\lim f$	$\lim g$	$\lim \frac{f}{g}$
$L$	$L'\neq 0$	$\frac{L}{L'}$
Example: $\lim_{x\to 1}\frac{x^{2}+3}{2x+6}=\frac{4}{8}=\frac{1}{2}$
$L$	$\pm\infty$	$0$
Example: $\lim_{x\to+\infty}\frac{9}{2x+1}=0$ : this is the line that gives the horizontal asymptote $y=0$ .
$\pm\infty$	$L'\neq 0$	$\pm\infty$ , sign rule
Example: $\lim_{x\to 3^{+}}\frac{\frac{1}{x-3}}{x-8}=\frac{+\infty}{-5}=-\infty$
$L\neq 0$	$0^{+}$ or $0^{-}$	$\pm\infty$ , sign rule
Example: $\lim_{x\to 3^{+}}\frac{x+1}{x-3}=+\infty$ because $x-3$ is $0^{+}$ , whereas $\lim_{x\to 3^{-}}\frac{x+1}{x-3}=-\infty$ because $x-3$ is $0^{-}$ . So the limit at $3$ does not exist, and yet $x=3$ is indeed a vertical asymptote.
$L>0$	$0^{+}$ on both sides	$+\infty$ on both sides
Example: $\lim_{x\to 3}\frac{x+1}{(x-3)^{2}}=+\infty$ : a square in the denominator stays positive on both sides, so the two one-sided limits are equal and the limit exists, as an infinite limit.
$0$	$0$	$\frac{0}{0}$ settles nothing
Example: $\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\frac{1}{4}$ , after multiplying by the conjugate $\sqrt{x}+2$ . Same form, other result: $\lim_{x\to 4}\frac{x-4}{\sqrt{x}-2}=\lim_{x\to 4}\left(\sqrt{x}+2\right)=4$ : the same form $\frac{0}{0}$ , result $4$ instead of $\frac{1}{4}$ , just by swapping top and bottom. What to do: Factor out $(x-a)$ on top and bottom if there are polynomials, multiply by the conjugate if there is a square root, combine over a common denominator if there are nested fractions.
$\pm\infty$	$\pm\infty$	$\frac{\infty}{\infty}$ settles nothing
Example: $\lim_{x\to+\infty}\frac{2x^{2}-1}{5x^{2}+x}=\frac{2}{5}$ , after factoring $x^{2}$ out of the top and the bottom. Same form, other result: $\lim_{x\to+\infty}\frac{2x^{3}-1}{5x^{2}+x}=+\infty$ : the same form $\frac{\infty}{\infty}$ , and a single degree of difference changes everything. What to do: Factor out the highest power of the denominator, or compare the degrees with the table below.

The cell $\frac{L}{0}$ does not exist in this table: as long as the zero is not written $0^{+}$ or $0^{-}$ , line 4 is not filled in and the limit has not been computed. The sign is read from a sign chart of the denominator, never from the calculator's graph.

Power: the limit of $f^{\,g}$

The table of the last three indeterminate forms, the ones people forget because they do not look like fractions. They all share one feature: the base tends to $1$ or to $0$ , which seems to decide everything, and decides nothing.

$\lim f$ (base)	$\lim g$ (exponent)	$\lim f^{\,g}$
$L>0$	$L'$	$L^{L'}$
Example: $\lim_{x\to 9}x^{\frac{1}{2}}=9^{\frac{1}{2}}=3$
$L>1$	$+\infty$	$+\infty$
Example: $\lim_{x\to+\infty}\left(2+\frac{1}{x}\right)^{x}=+\infty$ : a base that stays above $2$ , raised to a power that blows up.
$0<L<1$	$+\infty$	$0$
Example: $\lim_{x\to+\infty}\left(\frac{1}{2}\right)^{x}=0$ : from $x=10$ on, the value is already below $0.001$ .
$1$	$\pm\infty$	$1^{\infty}$ settles nothing
Example: $\lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^{x}=\mathrm{e}\approx 2.718$ . This is the very definition of the number $\mathrm{e}$ . Same form, other result: $\lim_{x\to+\infty}\left(1+\frac{2}{x}\right)^{x}=\mathrm{e}^{2}\approx 7.389$ : the same form $1^{\infty}$ , and that single added $2$ changes the answer completely. What to do: Let $y=f^{\,g}$ , take logarithms, $\ln y=g\ln f$ , compute the limit of $\ln y$ , then go back up with the exponential.
$+\infty$	$0$	$\infty^{0}$ settles nothing
Example: $\lim_{x\to+\infty}x^{\frac{1}{x}}=1$ : at $x=1000$ , the value is already about $1.0069$ . Same form, other result: $\lim_{x\to+\infty}\left(\mathrm{e}^{x}\right)^{\frac{1}{x}}=\mathrm{e}\approx 2.718$ : the same form $\infty^{0}$ , a different result. What to do: Same move: logarithm, limit of $g\ln f$ , then exponential.
$0^{+}$	$0$	$0^{0}$ settles nothing
Example: $\lim_{x\to 0^{+}}x^{x}=1$ : at $x=0.001$ , the value is about $0.993$ . Same form, other result: $\lim_{x\to 0^{+}}\left(\mathrm{e}^{-\frac{1}{x}}\right)^{x}=\mathrm{e}^{-1}\approx 0.368$ : the same form $0^{0}$ , a different result. What to do: Same move: logarithm, limit of $g\ln f$ , then exponential.

The three red lines all come down to $0\cdot\infty$ through the logarithm, and it is L'Hôpital's rule that settles them, once it has been taught. On the first test, what is expected is often more modest: recognizing that the form is indeterminate and not answering $1$ by reflex.

Quotient of polynomials at $\pm\infty$ : the limit is read from the degrees

This table is a shortcut for the case $\frac{\infty}{\infty}$ , and it gives the asymptote at the same time. On a test, it justifies the answer but does not replace factoring out, which is what the grader wants to see.

Degrees compared	What happens	Limit at $\pm\infty$
top $<$ bottom	the denominator crushes the numerator	$0$ , horizontal asymptote $y=0$
Example: $\lim_{x\to+\infty}\frac{3x+1}{x^{2}-5}=0$ : degree $1$ against degree $2$ .
top $=$ bottom	the two dominant terms balance	ratio of the leading coefficients
Example: $\lim_{x\to+\infty}\frac{4x^{2}-x}{2x^{2}+7}=\frac{4}{2}=2$ , so a horizontal asymptote $y=2$ .
top $=$ bottom $+\,1$	long division	$\pm\infty$ , with an OBLIQUE asymptote
Example: $\frac{x^{2}+1}{x}=x+\frac{1}{x}\to+\infty$ , and the line $y=x$ is an oblique asymptote since $\frac{1}{x}\to 0$ .
top $\geq$ bottom $+\,2$	nothing to recover	$\pm\infty$ , no asymptote
Example: $\lim_{x\to+\infty}\frac{x^{3}+1}{x}=+\infty$ : the quotient behaves like $x^{2}$ , with neither a horizontal nor an oblique asymptote.
with a square root, at $-\infty$	$\sqrt{x^{2}}=\|x\|=-x$	the sign is reversed
Example: $\lim_{x\to+\infty}\frac{\sqrt{x^{2}+1}}{x}=1$ but $\lim_{x\to-\infty}\frac{\sqrt{x^{2}+1}}{x}=-1$ : two horizontal asymptotes for a single curve, $y=1$ and $y=-1$ .

The last line is the trap most worth knowing in the whole chapter: as soon as a square root comes out of an $x^{2}$ , write $\sqrt{x^{2}}=|x|$ and treat the two infinities separately.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Confusing the limit with the value of the function

2 marks

What not to write

“ $f(2)$ does not exist, so $\lim_{x\to 2}f(x)$ does not exist.”

What to write

“ $f$ is not defined at $2$ , but $\lim_{x\to 2}f(x)=4$ : the limit never looks at the point itself.”

From both sides, the curve heads towards height

4

\lim_{x\to 2}f(x)=4

, even though the point

(2, 4)

is missing. Concluding “no

f(2)

, so no limit” is wrong.

Why: This is the most frequent error on the first test, and it makes everything that follows wrong: the derivative is precisely the limit of a quotient that does NOT exist at the point.

2. Treating $\infty$ as a number

the whole question

What not to write

“ $\frac{\infty}{\infty}=1$ ”, or “ $\infty-\infty=0$ ”.

What to write

“ $\frac{\infty}{\infty}$ is an indeterminate form: I factor out the dominant term, then conclude.”

The three quotients are of the form

\frac{\infty}{\infty}

and head for three different fates:

2

+\infty

and

0

. The form decides nothing; the dominant term settles it.

Why: $\infty$ is a behaviour, not a value. $\frac{2x}{x}\to 2$ , $\frac{x^{2}}{x}\to+\infty$ and $\frac{x}{x^{2}}\to 0$ all have the same form and three different answers: the form alone decides nothing.

3. Cancelling a factor without saying it is nonzero

1 mark for rigour

What not to write

$\frac{(x-2)(x+2)}{x-2}=x+2$

What to write

$\frac{(x-2)(x+2)}{x-2}=x+2$ for $x\neq 2$ , so $\lim_{x\to 2}=4$ .

Why: The two functions are not equal: they agree everywhere EXCEPT at $2$ . It is precisely because you are working with $x\neq 2$ that cancelling is allowed, and that is the sentence the grader is looking for.

4. The square root at $-\infty$

2 marks, and the missing asymptote

What not to write

$\lim_{x\to-\infty}\frac{\sqrt{x^{2}+1}}{x}=1$

What to write

At $-\infty$ , $\sqrt{x^{2}+1}=|x|\sqrt{1+\frac{1}{x^{2}}}=-x\sqrt{1+\frac{1}{x^{2}}}$ , so the limit is $-1$ .

-\infty

the graph of

\sqrt{x^{2}+1}

follows

y=-x

, not

y=x

: divided by

x

, which is negative, it therefore tends to

-1

. The root stays positive on both sides; it is the sign of

x

that changes.

Why: This function has TWO horizontal asymptotes, $y=1$ and $y=-1$ . A test asking for “all the asymptotes” penalizes half the answer.

5. The $\lim$ symbol that disappears halfway through the calculation

0.5 to 1 mark per paper

What not to write

$\lim_{x\to 3}(x^{2}-9)=x^{2}-9=0$

What to write

$\lim_{x\to 3}(x^{2}-9)=3^{2}-9=0$ , with the $\lim$ symbol written on EVERY line until the substitution has been made.

Why: Without the symbol, you are writing an equality between a function and a number, which is false. Graders flag it every time, and it is the easiest mark not to lose.

6. Concluding uniqueness from the Intermediate Value Theorem

1 mark, and the next question

What not to write

“By the IVT, there is EXACTLY ONE solution $c$ such that $f(c)=0$ .”

What to write

“By the IVT, there exists AT LEAST one $c$ such that $f(c)=0$ . Moreover, $f$ is strictly increasing on $[a,b]$ , so this $c$ is unique.”

Continuous on

[a,b]

, negative at

a

, positive at

b

: the IVT guarantees a zero, and this curve has THREE (in red). Uniqueness also requires strict monotonicity.

Why: The IVT gives existence, never uniqueness. Uniqueness comes from strict monotonicity, which is justified by the sign of the derivative: it is a second, separate sentence.

7. Applying $\frac{\sin x}{x}\to 1$ to an argument that is not the denominator

2 marks

What not to write

$\lim_{x\to 0}\frac{\sin 3x}{x}=1$

What to write

$\frac{\sin 3x}{x}=3\cdot\frac{\sin 3x}{3x}\to 3\cdot 1=3$ .

Both graphs have a hole at

x=0

, but not at the same height:

\frac{\sin 3x}{x}

arrives at

3

, not at

1

. The reference limit requires the same

3x

on top and on the bottom.

Why: The reference limit is about $\frac{\sin u}{u}$ with the SAME $u$ on top and on the bottom. You create that $u$ by multiplying and dividing by the coefficient, which leaves the factor in front.

8. Declaring a vertical asymptote as soon as the denominator vanishes

1 to 2 marks

What not to write

“ $x=2$ makes the denominator of $\frac{x^{2}-4}{x-2}$ zero, so $x=2$ is a vertical asymptote.”

What to write

“The numerator also vanishes at $2$ ; after cancelling, the limit is $4$ : there is a HOLE at $x=2$ , not an asymptote.”

The denominator of

\frac{x^{2}-4}{x-2}

vanishes at

x=2

and yet no branch heads off to infinity: the numerator vanishes at the same point, the factor

(x-2)

cancels, and all that remains is a hole at

(2, 4)

Why: A vertical asymptote requires an INFINITE limit, not a zero denominator. First check whether the numerator vanishes at the same point: that is what tells a hole from an asymptote.

9. Writing an $\infty$ without a sign when the form is $\frac{k}{0}$

1 to 2 marks

What not to write

“ $\lim_{x\to 2}\frac{x^{2}+4}{x-2}=\infty$ .”

What to write

“On the left, $x-2<0$ and the numerator equals $8>0$ , so the limit is $-\infty$ ; on the right it is $+\infty$ . The two differ: $\lim_{x\to 2}\frac{x^{2}+4}{x-2}$ DOES NOT EXIST, and $x=2$ is still a vertical asymptote.”

Form

\frac{8}{0}

: the left branch dives towards

-\infty

, the right branch climbs towards

+\infty

. A vertical asymptote, yes; a limit at

2

, no. An “

\infty

” without a sign says none of this.

Why: The form $\frac{k}{0}$ with $k\neq 0$ is NOT an indeterminate form: there is nothing to factor, and hunting for a common factor wastes five minutes. But it is not an answer either: it forces you to study the SIGN on each side, and the two sides can contradict each other. Compare with $\lim_{x\to 2}\frac{x^{2}+4}{(x-2)^{2}}=+\infty$ , where the square makes the denominator positive on both sides and the limit exists, as an infinite limit. Finally, the symbol $\infty$ without a sign means nothing in differential calculus: always write $+\infty$ or $-\infty$ .

10. Distributing the square root over a sum

the whole question

What not to write

$\sqrt{4x^{2}-5x}=2x-\sqrt{5x}$ , then a conjugate to get out of it.

What to write

$\sqrt{4x^{2}-5x}=|x|\sqrt{4-\frac{5}{x}}$ : factor $x^{2}$ out under the root, and at $-\infty$ , $|x|=-x$ .

The right triangle with legs

3

and

4

has a hypotenuse of

\sqrt{9+16}=5

, shorter than the path

3+4=7

: the root of a sum is not the sum of the roots.

Why: The square root distributes over neither a sum nor a difference: $\sqrt{9+16}=5$ , not $3+4$ . And without a difference of roots, the conjugate simplifies nothing; it adds a line of calculation where the error hides.

11. Looking for specific values when the question asks about the jump

the whole question

What not to write

“The one-sided limits are equal for $k=-1$ or $k=3$ , so there is a jump for $k=-1$ or $k=3$ .”

What to write

“The one-sided limits are equal only for $k=-1$ and $k=3$ , so they differ, and there is a jump, for every $k\in\mathbb{R}\setminus\{-1, 3\}$ .”

Why: A jump requires DIFFERENT one-sided limits: it is the complement of the values that make the limit exist. And among those values, the one where $f(a)$ differs from the limit gives a removable discontinuity, not continuity.

12. Keeping the direction of the inequalities when multiplying by a negative

the marks for the justification

What not to write

For $x\to 0^{-}$ : $-x\le x\cos\frac{1}{x}\le x$ .

What to write

For $x<0$ , multiplying by $x$ reverses the inequalities: $x\le x\cos\frac{1}{x}\le -x$ , and both bounds tend to 0.

For

x<0

, it is

y=-x

that lies ABOVE:

x\le x\cos\frac{1}{x}\le -x

. Both bounds tend to

0

, and so does the curve trapped between them.

Why: The wrong bound gives itself away: at $x=-0.1$ , it claims $0.1\le\dots\le -0.1$ . The limit is still 0, so the error does not show in the answer, but the Squeeze Theorem requires a true double inequality, and that is what the grader reads.

Which method to choose

Which method resolves the indeterminate form

The method does not depend on the chapter: it depends on the FORM of the expression, which you read before computing anything.

If form $\frac{0}{0}$ with polynomials → factor top and bottom, the common factor is ALWAYS $(x-a)$ , then cancel

Example: $\lim_{x\to 2}\frac{x^{2}-x-2}{x-2}=\lim_{x\to 2}(x+1)=3$

For a polynomial of degree 3 or more, synthetic division by $(x-a)$ is faster than factoring by inspection.
If form $\frac{0}{0}$ with a square root → multiply top and bottom by the conjugate expression, which turns the root into a difference of squares

Example: $\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\frac{x}{x(\sqrt{x+4}+2)}\to\frac{1}{4}$
If form $\frac{0}{0}$ with nested fractions → combine the compound numerator over a common denominator, then simplify the stacked fraction

Example: $\frac{\frac{1}{x}-\frac{1}{3}}{x-3}=\frac{3-x}{3x(x-3)}=-\frac{1}{3x}\to-\frac{1}{9}$
If form $\frac{\infty}{\infty}$ , rational function → compare the degrees, or factor out the highest power present in the denominator

Example: $\frac{3x^{2}-1}{5x^{2}+x}\to\frac{3}{5}$
If form $\infty-\infty$ → factor out the dominant term, or multiply by the conjugate if there is a square root

Example: $\sqrt{x^{2}+x}-x=\frac{x}{\sqrt{x^{2}+x}+x}\to\frac{1}{2}$ at $+\infty$
If form $\frac{0}{0}$ with sine or cosine → build $\frac{\sin u}{u}$ by multiplying and dividing by what is needed, or bound the expression and apply the Squeeze Theorem

Example: $\frac{1-\cos x}{x}=\frac{1-\cos x}{x^{2}}\cdot x\to\frac{1}{2}\cdot 0=0$

If no branch applies, that is the signal that there is NO indeterminate form: substitute, and the answer is immediate. In 201-SN2-RE, L'Hôpital's rule is not yet available at this point, and using it ahead of time is not accepted.

Which theorem for which question

The Part B questions all look alike. The verb in the question points to the theorem.

If the question says “show that there exists a solution” or “that the equation has a root” → Intermediate Value Theorem, with both of its hypotheses checked explicitly

Example: show that $x^{3}+x-1=0$ has a root in $[0,1]$
If the question adds “a unique solution” → IVT for existence, THEN strict monotonicity for uniqueness, in two separate sentences
If the question gives a double inequality, or a bounded expression multiplied by an expression that tends to $0$ → Squeeze Theorem

Example: $-x^{2}\le x^{2}\sin\frac{1}{x}\le x^{2}$ , so the limit at $0$ is $0$
If the question asks you to “find $a$ so that $f$ is continuous” → the three conditions for continuity at the junction point, turned into equations

Example: see the worked example taken apart below
If the question asks you to justify that a maximum is ATTAINED → Extreme Value Theorem, which requires a function continuous on a CLOSED and BOUNDED interval

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

The Intermediate Value Theorem, written up for the grader

When to use it: Whenever the question asks you to show that an equation has a solution on an interval, without computing it.

1 Name the function and the interval: “Let $f(x)=x^{3}+x-1$ on $[0,1]$ .”
2 Justify continuity by its REASON, never by “it is obvious”: “ $f$ is a polynomial, hence continuous on $\mathbb{R}$ , in particular on $[0,1]$ .”
3 Compute the two values at the endpoints and write them down: “ $f(0)=-1$ and $f(1)=1$ .”
4 Note that the target lies between the two: “ $0$ lies between $f(0)=-1$ and $f(1)=1$ .”
5 Conclude with the model sentence, naming the theorem.

Concluding sentence

Since $f$ is continuous on $[0,1]$ and $0$ lies between $f(0)$ and $f(1)$ , the Intermediate Value Theorem guarantees that there exists at least one real number $c\in\,(0,1)$ such that $f(c)=0$ .

The trap: Skipping step 2. A paper that computes $f(0)$ and $f(1)$ and then concludes directly loses the continuity mark, and continuity is the most important hypothesis of the theorem.

Marking: Usually 1 mark for justified continuity, 1 mark for the endpoint values and the inequality, 1 mark for the conclusion naming the theorem.

Finding a parameter that makes a function continuous

When to use it: A piecewise-defined function, with one or two parameters to find.

1 Locate the only points where continuity can fail: the JUNCTION points. Elsewhere, say why each piece is continuous (a polynomial, a quotient whose denominator does not vanish, a root on its domain).
2 Write the left-hand limit at the junction point, resolving the indeterminate form if the left piece has one.
3 Write the right-hand limit, then the value $f(a)$ , paying attention to the piece that REALLY contains the point (check whether the inequality is strict or not).
4 Set the three quantities equal, which gives one equation per junction point.
5 Solve the system, then check by substituting the values found.

Concluding sentence

For $a=3$ and $b=-3$ , the left-hand and right-hand limits agree with the value of the function at each junction point: $f$ is therefore continuous on $\mathbb{R}$ .

The trap: Looking for the value $f(a)$ in the wrong piece. An $x<2$ on one side and a $2\le x$ on the other mean that $f(2)$ belongs to the SECOND piece, the one with the non-strict inequality.

Marking: One correct equation per junction point, then the solution. The written conclusion earns the last mark.

Proving a limit with the formal epsilon-delta definition

When to use it: A rigour question, almost always on a simple linear or quadratic function, at the end of the test.

1 Write what must be proved: “for every $\varepsilon>0$ , there exists $\delta>0$ such that $0<|x-a|<\delta\Rightarrow|f(x)-L|<\varepsilon$ .”
2 Work on SCRAP PAPER, starting from $|f(x)-L|$ and bounding it by an expression in $|x-a|$ .
3 Isolate $|x-a|$ to read off the $\delta$ to choose, for example $\delta=\frac{\varepsilon}{3}$ .
4 Write it up in the FORWARD direction: “Let $\varepsilon>0$ . Take $\delta=\dots$ If $0<|x-a|<\delta$ , then $|f(x)-L|=\dots<\varepsilon$ .”
5 Conclude that the definition is satisfied, so that the limit is $L$ .

Concluding sentence

For every $\varepsilon>0$ , the choice $\delta=\frac{\varepsilon}{3}$ works, so $\lim_{x\to a}f(x)=L$ by definition.

The trap: Handing in the scratch work as the proof. The search for $\delta$ goes back from the conclusion to the hypothesis; the proof must go down from the hypothesis to the conclusion. The grader marks the version written in the right direction.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Make the calculator confirm the answer before you hand it in

On a TI, enter the function in Y1, then use TABLE with a step of $0.001$ around the point: the column of values converges to the announced limit. On a Casio, the TABLE menu with Start and Step does the same job.

A limit announced as $4$ with a table showing $3.998$ then $3.999$ is confirmed; a table showing $7.9$ says the factoring is wrong.
Test both sides, not just one

Evaluate at $a-0.001$ AND at $a+0.001$ . Two values that move away from each other signal a jump or an asymptote, hence a limit that does not exist.

This is the check that catches a forgotten $\sqrt{x^{2}}=|x|$ : at $-10^{6}$ the calculator shows $-1$ , not $1$ .
Mentally replace by the dominant terms

At $\pm\infty$ , keep only the highest-degree term on top and on the bottom: the ratio you get must match your result.

$\frac{3x^{2}-1}{5x^{2}+x}$ reads as $\frac{3x^{2}}{5x^{2}}=\frac{3}{5}$ in three seconds.
Check RADIAN mode before any trigonometric limit

$\lim_{x\to 0}\frac{\sin x}{x}=1$ is true only in radians. In degrees, the calculator shows $0.0174\ldots$ , which casts doubt on a calculation that is in fact correct.

The number $0.01745$ on the screen is the signal: it is $\frac{\pi}{180}$ , and the calculator is in degrees.
Reread the question asked

“Does the limit exist?” is answered yes or no with a justification, “compute the limit” with a number, “is the function continuous?” with the three conditions. A correct answer to the wrong question earns nothing.

On a graph, $f(a)$ was asked and a limit was computed instead: the answer is correct, the mark is zero.

The typical problem, taken apart

A piecewise function with two parameters, as on the first test

Let $f$ be defined on $\mathbb{R}$ by: $f(x)=\frac{x^{2}-x-2}{x-2}$ if $x<2$ , $f(x)=ax+b$ if $2\le x<5$ , and $f(x)=2x+2$ if $x\ge 5$ .

Find the real numbers $a$ and $b$ so that $f$ is continuous on $\mathbb{R}$ .

Step 1

On $(-\infty,2)$ , $f$ is a quotient of polynomials whose denominator does not vanish; on $(2,5)$ and $(5,+\infty)$ , $f$ is linear. The only points to examine are $x=2$ and $x=5$ .

Why

Start by saying WHERE the problem can arise. Without this sentence, the paper looks as if it is treating two points at random, and the grader takes off the justification mark.

Step 2

$\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{-}}\frac{x^{2}-x-2}{x-2}$ is of the form $\frac{0}{0}$ ; now $x^{2}-x-2=(x-2)(x+1)$ , so for $x\neq 2$ the expression equals $x+1$ and the limit is $3$ .

Why

Indeterminate form announced, factoring, mention of $x\neq 2$ which allows the cancellation: the three moves of the chapter in one line. This is where the rigour mark is earned.

Step 3

The inequality $2\le x$ is NON-STRICT, so $f(2)=2a+b$ , and since the middle piece is linear, $\lim_{x\to 2^{+}}f(x)=2a+b$ as well. Continuity at $2$ requires $2a+b=3$ .

Why

The trap with piecewise functions is to look for $f(2)$ in the left piece. Checking which of the two inequalities is non-strict settles the question once and for all.

Step 4

At $5$ : $\lim_{x\to 5^{-}}f(x)=5a+b$ , while $f(5)=2(5)+2=12$ since $x\ge 5$ belongs to the third piece. Continuity at $5$ requires $5a+b=12$ .

Why

The same reading as in the previous step, but the non-strict inequality has changed sides: the value at the point now comes from the RIGHT piece.

Step 5

The system $\begin{cases}2a+b=3\\5a+b=12\end{cases}$ gives, by subtraction, $3a=9$ so $a=3$ , then $b=3-2(3)=-3$ .

Why

Subtracting the two equations eliminates $b$ immediately: no need to isolate and substitute, the system is already set up for elimination.

Step 6

Check: with $a=3$ and $b=-3$ , the middle piece is $3x-3$ . At $2$ : $3(2)-3=3$ , which is indeed the left-hand limit. At $5$ : $3(5)-3=12$ , which is indeed $f(5)$ .

Why

Two ten-second substitutions. They catch the most common sign error in the chapter, the one that gives $b=3$ instead of $b=-3$ .

The conclusion, written out

For $a=3$ and $b=-3$ , the left-hand and right-hand limits agree with the value of $f$ at $2$ and at $5$ : $f$ is continuous at these two points, and therefore continuous on all of $\mathbb{R}$ .

The classic mistake on this problem: Substituting $x=2$ into the first piece, getting $\frac{0}{0}$ and concluding that $f$ cannot be continuous. The form $\frac{0}{0}$ says nothing: it calls for factoring, after which the limit exists and equals $3$ .

Learn by heart

• $\lim_{x\to a}f(x)$ exists if and only if both one-sided limits exist and are EQUAL; $f(a)$ plays no part.
• Continuity at $a$ : $f(a)$ exists, the limit exists, and the two are equal. Three conditions, never two.
• $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2}$ , in radians, with the same argument on top and on the bottom.
• $\sqrt{x^{2}}=|x|$ : at $-\infty$ this equals $-x$ , hence the possibility of two horizontal asymptotes.
• Rational function at $\pm\infty$ : compare the degrees. Lower on top, limit $0$ ; equal degrees, ratio of the leading coefficients; one degree higher, oblique asymptote.
• The Intermediate Value Theorem gives EXISTENCE. Uniqueness comes from strict monotonicity, in a separate sentence.
• An indeterminate form is not an answer: it is an order to transform the expression.
• Continuity on an interval: at a CLOSED endpoint, only one-sided continuity on the side of the interval is required. A jump at the right endpoint of an interval does not stop the function from being continuous on it.
• REMOVABLE discontinuity, the limit exists: a hole, whether $f(a)$ does not exist or differs from the limit. NON-REMOVABLE discontinuity, the limit does not exist: a jump or an infinite discontinuity.

Frequently asked questions

What is the difference between the limit of a function at a point and its value at that point?

The limit describes what the function does around the point; the value is what it equals at the point. The two can both exist and be different, and the limit can perfectly well exist where the function is not defined. This is exactly the distinction that every question on continuity tests.

How do you resolve a 0 over 0 indeterminate form?

You transform the expression, you do not compute. With polynomials, factor and cancel the factor that vanishes. With a square root, multiply by the conjugate expression. With nested fractions, combine over a common denominator. An indeterminate form is not an answer; it is an order to change the form.

How do you show that a function is continuous at a point?

Three conditions, in this order, and all three written on your paper: the function is defined at the point, the limit exists at that point, and these two numbers are equal. For a piecewise-defined function, compute the two one-sided limits at the junction point, then set them equal to the required value.

When do you use the Intermediate Value Theorem?

As soon as the question asks you to show that an equation has a solution on an interval. Check continuity on the closed interval, compute the values at the two endpoints, note the change of sign, then conclude that a solution exists. Uniqueness never comes from this theorem: it is justified separately, by strict monotonicity.

Why can a square root give two horizontal asymptotes?

Because the square root of x squared equals the absolute value of x: that gives x at plus infinity, but minus x at minus infinity. The sign changes with the side, so the two limits at infinity can be opposite. It is the classic trap for functions with a square root in the numerator or the denominator.

Can a function be continuous on one interval and discontinuous on another containing the same point?

Yes, and that is exactly what the first test checks. At a closed endpoint, continuity is required on one side only, the side where the interval lies. A jump at the right endpoint of an interval therefore does not stop the function from being continuous on that interval, while the same jump in the middle of a larger interval makes it discontinuous, and an interval that is open at that point makes it continuous again. The question is never whether the function is continuous, but whether it is continuous on that particular interval.

Practise it

Corrected exercises: Limits and continuity, 201-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

17 corrected exercises
170 points
225 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-calc1-limits-continuity. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: limits and continuity (201-SN2-RE)

The essentials

Three different questions at a single point

The seven indeterminate forms, and the ones that are not

Continuity at aaa: three conditions, in this order

Continuity ON AN INTERVAL: the endpoint changes the answer

Limits to recognize at a glance

The rules in table form

Sum: the limit of f+gf+gf+g

Product: the limit of f×gf\times gf×g

Quotient: the limit of fg\frac{f}{g}gf​

Power: the limit of f gf^{\,g}fg

Quotient of polynomials at ±∞\pm\infty±∞: the limit is read from the degrees

The mistakes that cost marks

1. Confusing the limit with the value of the function

2. Treating ∞\infty∞ as a number

3. Cancelling a factor without saying it is nonzero

4. The square root at −∞-\infty−∞

5. The lim⁡\limlim symbol that disappears halfway through the calculation

6. Concluding uniqueness from the Intermediate Value Theorem

7. Applying sin⁡xx→1\frac{\sin x}{x}\to 1xsinx​→1 to an argument that is not the denominator

8. Declaring a vertical asymptote as soon as the denominator vanishes

9. Writing an ∞\infty∞ without a sign when the form is k0\frac{k}{0}0k​

10. Distributing the square root over a sum

11. Looking for specific values when the question asks about the jump

12. Keeping the direction of the inequalities when multiplying by a negative

Which method to choose

Which method resolves the indeterminate form

Which theorem for which question

How the answer is expected to be written

The Intermediate Value Theorem, written up for the grader

Finding a parameter that makes a function continuous

Proving a limit with the formal epsilon-delta definition

Check before you hand in

The typical problem, taken apart

A piecewise function with two parameters, as on the first test

Learn by heart

Frequently asked questions

What is the difference between the limit of a function at a point and its value at that point?

How do you resolve a 0 over 0 indeterminate form?

How do you show that a function is continuous at a point?

When do you use the Intermediate Value Theorem?

Why can a square root give two horizontal asymptotes?

Can a function be continuous on one interval and discontinuous on another containing the same point?

Corrected exercises: Limits and continuity, 201-SN2-RE

See also

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Continuity at $a$ : three conditions, in this order

Sum: the limit of $f+g$

Product: the limit of $f\times g$

Quotient: the limit of $\frac{f}{g}$

Power: the limit of $f^{\,g}$

Quotient of polynomials at $\pm\infty$ : the limit is read from the degrees

2. Treating $\infty$ as a number

4. The square root at $-\infty$

5. The $\lim$ symbol that disappears halfway through the calculation

7. Applying $\frac{\sin x}{x}\to 1$ to an argument that is not the denominator

9. Writing an $\infty$ without a sign when the form is $\frac{k}{0}$