Exercise 1: Limits at infinity: divide EVERY term by the dominant power

Thomas's basic facts: $\lim_{x\to\pm\infty} k = k$ and $\lim_{x\to\pm\infty} \frac{1}{x} = 0$ , so $\frac{c}{x^r} \to 0$ for every $r > 0$ (at $-\infty$ whenever $x^r$ is defined). For a quotient, divide the numerator AND the denominator, every term of each, by the highest power of $x$ that appears in the DENOMINATOR, then let every $\frac{c}{x^r}$ go to $0$ .

The same method works with negative and fractional exponents, once every term is written as a power of $x$ : $\sqrt{x} = x^{1/2}$ , $x^{-3} = \frac{1}{x^3}$ , and $\frac{x^a}{x^b} = x^{a-b}$ . The figure shows $f(x) = \frac{2x^2 - 8}{x^2 + 3}$ .

a) Read on the figure $\lim_{x\to\infty} f(x)$ and $\lim_{x\to -\infty} f(x)$ , then prove both. Does the graph of $f$ ever reach its horizontal asymptote?
b) Find $\lim_{x\to\infty} \frac{5 - 4x^3}{2x^2 + x}$ and $\lim_{x\to -\infty} \frac{5 - 4x^3}{2x^2 + x}$ .
c) Find $\lim_{x\to\infty} \frac{6x^{-1} + x^{-3}}{2x^{-2} - x^{-1}}$ .
d) Find $\lim_{x\to\infty} \frac{4x - x^{2/3} + 1}{3x + \sqrt{x}}$ .
e) Find $\lim_{x\to\infty} \frac{x^{3/2} - 7x}{2x + 9}$ and $\lim_{x\to\infty} \frac{3x + 1}{x^{3/2} + 2}$ .

Type your answers, the page tells you right or wrong 0/9

a)

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

\lim_{x\to -\infty} f(x)

Does the graph ever reach

y = 2

?

b)

\lim_{x\to\infty}

\lim_{x\to -\infty}

c)

The limit

d)

The limit

e)

First limit

Second limit

Show the solution

Answers

a) $2$ at both ends; $f(x) - 2 = -\frac{14}{x^2 + 3} < 0$ , so the graph never reaches $y = 2$ and approaches it from below.
b) $-\infty$ as $x \to \infty$ , $+\infty$ as $x \to -\infty$ : no horizontal asymptote.
c) $-6$
d) $\frac{4}{3}$
e) $+\infty$ and $0$

a) The figure shows the curve flattening onto the line $y = 2$ at both ends. Proof: the highest power of the denominator is $x^2$ , and dividing EVERY term by it gives $f(x) = \frac{2 - \frac{8}{x^2}}{1 + \frac{3}{x^2}} \to \frac{2 - 0}{1 + 0} = 2$ as $x \to \infty$ and as $x \to -\infty$ . So $y = 2$ is the horizontal asymptote at both ends. To see whether the graph reaches it, compute the gap on a common denominator: $f(x) - 2 = \frac{2x^2 - 8 - 2(x^2 + 3)}{x^2 + 3} = \frac{2x^2 - 8 - 2x^2 - 6}{x^2 + 3} = -\frac{14}{x^2 + 3}$ . The $-2$ must multiply BOTH terms of $x^2 + 3$ : writing $-2x^2 + 6$ gives a gap of $-2$ and a wrong conclusion. The gap is negative for every $x$ , so the graph stays strictly below $y = 2$ and never reaches it.

b) Divide every term by $x^2$ : $\frac{5 - 4x^3}{2x^2 + x} = \frac{\frac{5}{x^2} - 4x}{2 + \frac{1}{x}}$ . The denominator tends to $2$ . As $x \to \infty$ , $-4x \to -\infty$ , so the quotient tends to $-\infty$ ; as $x \to -\infty$ , $-4x \to +\infty$ , so the quotient tends to $+\infty$ . The shortcut, and the check, is the quotient of the dominant terms: $\frac{-4x^3}{2x^2} = -2x$ , which is odd, so the two ends go opposite ways. Degree $3$ over degree $2$ : no horizontal asymptote.

c) Rewrite the exponents before choosing anything: $\frac{6x^{-1} + x^{-3}}{2x^{-2} - x^{-1}} = \frac{\frac{6}{x} + \frac{1}{x^3}}{\frac{2}{x^2} - \frac{1}{x}}$ . Clear the small fractions by multiplying the numerator and the denominator by $x^3$ : $\frac{6x^2 + 1}{2x - x^2}$ . Now divide by $x^2$ : $\frac{6 + \frac{1}{x^2}}{\frac{2}{x} - 1} \to \frac{6}{-1} = -6$ . The trap is to treat $x^{-3}$ as the dominant term because $3 > 1$ : at infinity, $x^{-3} = \frac{1}{x^3}$ is the SMALLEST term, and $x^{-1}$ the largest. Pairing the coefficients as they are written, $\frac{6}{2} = 3$ , is the other classic loss.

d) As $x \to \infty$ , $x$ is larger than $\sqrt{x} = x^{1/2}$ , so the dominant power of the denominator is $x$ . Divide every term by $x$ , subtracting exponents: $\frac{x^{2/3}}{x} = x^{2/3 - 1} = x^{-1/3} = \frac{1}{\sqrt[3]{x}}$ and $\frac{\sqrt{x}}{x} = x^{-1/2} = \frac{1}{\sqrt{x}}$ . Then $\frac{4 - x^{-1/3} + \frac{1}{x}}{3 + x^{-1/2}} \to \frac{4}{3}$ . The exponent rule is $\frac{x^a}{x^b} = x^{a-b}$ , not $x^{a/b}$ : $\frac{x^{2/3}}{x}$ is not $x^{2/3}$ .

e) First limit: the dominant power of the denominator is $x$ ; dividing, $\frac{x^{1/2} - 7}{2 + \frac{9}{x}}$ , whose numerator tends to $\infty$ because $\sqrt{x} \to \infty$ . The limit is $+\infty$ : the numerator's largest exponent, $\frac{3}{2}$ , beats the denominator's, $1$ . Second limit: now the denominator's dominant power is $x^{3/2}$ : $\frac{3x^{-1/2} + x^{-3/2}}{1 + 2x^{-3/2}} \to \frac{0}{1} = 0$ . The rule of degrees for rational functions survives with fractional exponents: compare the largest exponent on top with the largest exponent below.

Tick the exercises you have done or want to review: a free account, no password, keeps your ticks from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

Exercise 2: Square roots at infinity: the root of x squared is |x|

To divide a square root by a power of $x$ , factor $x^2$ out of the root: $\sqrt{9x^2 + 4} = \sqrt{x^2\left(9 + \frac{4}{x^2}\right)} = \sqrt{x^2}\,\sqrt{9 + \frac{4}{x^2}} = |x|\sqrt{9 + \frac{4}{x^2}}$ . For $x > 0$ , $|x| = x$ ; for $x < 0$ , $|x| = -x$ , and a minus sign appears that decides the limit at $-\infty$ .

A cube root behaves differently: $\sqrt[3]{x^3} = x$ for every real $x$ , with no absolute value. The figure shows $g(x) = \frac{\sqrt{9x^2 + 4}}{2x - 5}$ .

a) Find $\lim_{x\to\infty} g(x)$ .
b) Find $\lim_{x\to -\infty} g(x)$ . How many horizontal asymptotes does the graph of $g$ have?
c) Find $\lim_{x\to (5/2)^-} g(x)$ and $\lim_{x\to (5/2)^+} g(x)$ . Can the graph of $g$ have a hole instead?
d) Find $\lim_{x\to\infty} \left(\sqrt{x^2 + 3x} - \sqrt{x^2 - x}\right)$ and $\lim_{x\to -\infty} \left(\sqrt{x^2 + 3x} - \sqrt{x^2 - x}\right)$ .
e) Find $\lim_{x\to\infty} \frac{\sqrt[3]{8x^3 - x^2}}{x + 1}$ and $\lim_{x\to -\infty} \frac{\sqrt[3]{8x^3 - x^2}}{x + 1}$ .

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

a)

The limit

b)

The limit

Number of horizontal asymptotes

c)

\lim_{x\to (5/2)^-} g(x)

\lim_{x\to (5/2)^+} g(x)

Can there be a hole at

x = 5/2

?

d)

\lim_{x\to\infty}

\lim_{x\to -\infty}

e)

\lim_{x\to\infty}

\lim_{x\to -\infty}

Show the solution

Answers

a) $\frac{3}{2}$
b) $-\frac{3}{2}$ ; two horizontal asymptotes, $y = \frac{3}{2}$ on the right and $y = -\frac{3}{2}$ on the left.
c) $-\infty$ and $+\infty$ ; no hole is possible, since $9x^2 + 4 \ge 4$ never vanishes.
d) $2$ and $-2$
e) $2$ at both ends: one horizontal asymptote, $y = 2$ .

a) For $x > 0$ , $\sqrt{9x^2 + 4} = x\sqrt{9 + \frac{4}{x^2}}$ , and $2x - 5 = x\left(2 - \frac{5}{x}\right)$ . Cancel $x$ : $g(x) = \frac{\sqrt{9 + \frac{4}{x^2}}}{2 - \frac{5}{x}} \to \frac{\sqrt 9}{2} = \frac{3}{2}$ . The root of the limit is the limit of the root, since $\sqrt{\ }$ is continuous at $9$ .

b) For $x < 0$ , $\sqrt{x^2} = |x| = -x$ , so $\sqrt{9x^2 + 4} = -x\sqrt{9 + \frac{4}{x^2}}$ . Then $g(x) = \frac{-x\sqrt{9 + \frac{4}{x^2}}}{x\left(2 - \frac{5}{x}\right)} = -\frac{\sqrt{9 + \frac{4}{x^2}}}{2 - \frac{5}{x}} \to -\frac{3}{2}$ . The graph has TWO horizontal asymptotes, $y = \frac{3}{2}$ as $x \to \infty$ and $y = -\frac{3}{2}$ as $x \to -\infty$ , as the figure shows. Sanity check before any algebra: at $x = -100$ the root is positive and $2x - 5$ is negative, so $g < 0$ , and an answer of $+\frac{3}{2}$ is impossible. Writing $\sqrt{x^2} = x$ is the algebra slip that costs this whole limit.

c) As $x \to \frac{5}{2}$ , the numerator tends to $\sqrt{9 \cdot \frac{25}{4} + 4} = \sqrt{\frac{241}{4}} \approx 7.76 > 0$ , and $2x - 5 \to 0$ : through negative values on the left ( $0^-$ ), positive on the right ( $0^+$ ). So $\lim_{x\to (5/2)^-} g(x) = -\infty$ and $\lim_{x\to (5/2)^+} g(x) = +\infty$ , and $x = \frac{5}{2}$ is a vertical asymptote. A hole would need the numerator to tend to $0$ too, and $9x^2 + 4 \ge 4$ for every $x$ : impossible.

d) Both roots tend to $+\infty$ at each end: the form $\infty - \infty$ . Multiply and divide by the conjugate: $\sqrt{x^2 + 3x} - \sqrt{x^2 - x} = \frac{(x^2 + 3x) - (x^2 - x)}{\sqrt{x^2 + 3x} + \sqrt{x^2 - x}} = \frac{4x}{\sqrt{x^2 + 3x} + \sqrt{x^2 - x}}$ . The parentheses matter: subtracting the WHOLE of $x^2 - x$ gives $+x$ , hence $4x$ , not $2x$ . For $x > 0$ , each root is $x\sqrt{\cdots}$ : $\frac{4}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 - \frac{1}{x}}} \to \frac{4}{2} = 2$ . For $x < 0$ , each root is $-x\sqrt{\cdots}$ , so the denominator is $-x\left(\sqrt{1 + \frac{3}{x}} + \sqrt{1 - \frac{1}{x}}\right)$ and the quotient is $\frac{-4}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 - \frac{1}{x}}} \to -2$ . Calculator check at $x = 100$ : $\sqrt{10300} - \sqrt{9900} \approx 101.489 - 99.499 = 1.990$ .

e) $\sqrt[3]{8x^3 - x^2} = \sqrt[3]{x^3\left(8 - \frac{1}{x}\right)} = x\sqrt[3]{8 - \frac{1}{x}}$ , and this holds for EVERY $x \ne 0$ , negative or positive, because $\sqrt[3]{x^3} = x$ . With $x + 1 = x\left(1 + \frac{1}{x}\right)$ , the quotient is $\frac{\sqrt[3]{8 - \frac{1}{x}}}{1 + \frac{1}{x}} \to \sqrt[3]{8} = 2$ at both ends. One horizontal asymptote, $y = 2$ . Contrast with b): an EVEN root turns $x$ into $|x|$ and can split the two ends; an ODD root keeps the sign of $x$ and does not.

Exercise 3: Infinite limits: factor the denominator, then read the sign on each side

The line $x = a$ is a vertical asymptote of $y = f(x)$ when $\lim_{x\to a^+} f(x) = \pm\infty$ or $\lim_{x\to a^-} f(x) = \pm\infty$ . For a quotient whose numerator tends to $L \ne 0$ and whose denominator tends to $0$ , the size is infinite and the SIGN on each side is the whole question.

To read those signs, factor the denominator completely first: difference of squares, trinomial, difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ . When the numerator ALSO tends to $0$ , factor it too and cancel before deciding. The figure shows $f(x) = \frac{x + 1}{x^2 - 4x}$ .

a) Read on the figure the one-sided limits of $f$ at $0$ and at $4$ , and $\lim_{x\to\pm\infty} f(x)$ . Confirm the four one-sided limits from the factored formula.
b) Find $\lim_{x\to 2^-} \frac{3 - 2x}{x^2 - 4}$ and $\lim_{x\to 2^+} \frac{3 - 2x}{x^2 - 4}$ .
c) Let $k(x) = \frac{x^3 - 8}{x^2 - 4}$ . For $x = 2$ and $x = -2$ , decide whether the graph of $k$ has a vertical asymptote there, and give the limits.
d) Find $\lim_{x\to 0} \frac{1}{x^{2/3}}$ , $\lim_{x\to 0^-} \frac{1}{x^{1/3}}$ and $\lim_{x\to 0^+} \frac{1}{x^{1/3}}$ .
e) Find $\lim_{x\to 1^-} \frac{1}{\ln x}$ , $\lim_{x\to 1^+} \frac{1}{\ln x}$ , $\lim_{x\to \pi^-} \cot x$ and $\lim_{x\to \pi^+} \cot x$ .

Type your answers, the page tells you right or wrong 0/18

a)

\lim_{x\to 0^-} f(x)

\lim_{x\to 0^+} f(x)

\lim_{x\to 4^-} f(x)

\lim_{x\to 4^+} f(x)

\lim_{x\to\pm\infty} f(x)

b)

\lim_{x\to 2^-}

\lim_{x\to 2^+}

c)

Vertical asymptote at

x = 2

?

\lim_{x\to 2} k(x)

\lim_{x\to -2^-} k(x)

\lim_{x\to -2^+} k(x)

d)

\lim_{x\to 0} x^{-2/3}

\lim_{x\to 0^-} x^{-1/3}

\lim_{x\to 0^+} x^{-1/3}

e)

\lim_{x\to 1^-} \frac{1}{\ln x}

\lim_{x\to 1^+} \frac{1}{\ln x}

\lim_{x\to \pi^-} \cot x

\lim_{x\to \pi^+} \cot x

Show the solution

Answers

a) $+\infty$ as $x \to 0^-$ , $-\infty$ as $x \to 0^+$ ; $-\infty$ as $x \to 4^-$ , $+\infty$ as $x \to 4^+$ ; $0$ at both ends.
b) $+\infty$ from the left, $-\infty$ from the right.
c) $x = 2$ : a hole, $\lim_{x\to 2} k(x) = 3$ ; $x = -2$ : asymptote, $-\infty$ on the left and $+\infty$ on the right.
d) $+\infty$ ; $-\infty$ and $+\infty$ .
e) $-\infty$ , $+\infty$ ; $-\infty$ , $+\infty$ . Asymptotes $x = 1$ for $\frac{1}{\ln x}$ and $x = \pi$ for $\cot x$ .

a) On the figure: along $x = 0$ the curve climbs on the left and plunges on the right; along $x = 4$ it plunges on the left and climbs on the right; it flattens onto the $x$ -axis at both ends. From the formula, factor first: $f(x) = \frac{x + 1}{x(x - 4)}$ . Near $0$ the numerator tends to $1 > 0$ and $x - 4 \to -4 < 0$ . For $x < 0$ , $x(x - 4)$ is (negative)(negative), so the denominator tends to $0^+$ and $f \to +\infty$ ; for $0 < x < 4$ it is (positive)(negative), $0^-$ , and $f \to -\infty$ . Near $4$ the numerator tends to $5$ and $x \to 4$ , so the sign is that of $x - 4$ : $0^-$ on the left, $f \to -\infty$ ; $0^+$ on the right, $f \to +\infty$ . At $\pm\infty$ the degrees are $1$ over $2$ , so $f \to 0$ . Without the factorization, the sign of $x^2 - 4x$ near $0$ is a guess.

b) The numerator tends to $3 - 4 = -1$ . Factor the denominator: $x^2 - 4 = (x - 2)(x + 2)$ , with $x + 2 \to 4$ . As $x \to 2^-$ , $x - 2 \to 0^-$ , so the denominator tends to $0^-$ and the quotient is (negative)/(small negative): $+\infty$ . As $x \to 2^+$ the denominator tends to $0^+$ : $-\infty$ . A negative numerator reverses both signs; checking only the denominator gives the two answers backwards.

c) Factor both: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ (difference of cubes, check by expanding) and $x^2 - 4 = (x - 2)(x + 2)$ . For $x \ne 2$ , $k(x) = \frac{x^2 + 2x + 4}{x + 2}$ . At $x = 2$ the original formula is a form $\frac{0}{0}$ , and after cancelling $\lim_{x\to 2} k(x) = \frac{4 + 4 + 4}{4} = 3$ : a hole at $(2, 3)$ , no asymptote. At $x = -2$ the simplified numerator tends to $4 - 4 + 4 = 4 > 0$ and $x + 2 \to 0^-$ on the left, $0^+$ on the right: $-\infty$ then $+\infty$ , and $x = -2$ is a vertical asymptote. The factorizations $(x - 2)^3$ or $(x - 2)(x^2 + 4)$ are the usual slips: expand back to catch them.

d) $x^{2/3} = \left(\sqrt[3]{x}\right)^2$ is positive on both sides of $0$ and tends to $0$ , so $\frac{1}{x^{2/3}} \to +\infty$ from both sides and $\lim_{x\to 0} \frac{1}{x^{2/3}} = +\infty$ . But $x^{1/3} = \sqrt[3]{x}$ has the sign of $x$ : it tends to $0^-$ on the left and $0^+$ on the right, so $\frac{1}{x^{1/3}}$ tends to $-\infty$ on the left and $+\infty$ on the right, and $\lim_{x\to 0} \frac{1}{x^{1/3}}$ does not exist. Reading a fractional exponent means reading its root: an odd root keeps the sign, an even power removes it.

e) $\ln 1 = 0$ , $\ln x < 0$ for $0 < x < 1$ and $\ln x > 0$ for $x > 1$ . So $\frac{1}{\ln x} \to \frac{1}{0^-} = -\infty$ as $x \to 1^-$ and $+\infty$ as $x \to 1^+$ : $x = 1$ is a vertical asymptote. Next, $\cot x = \frac{\cos x}{\sin x}$ with $\cos x \to -1$ as $x \to \pi$ ; $\sin x > 0$ just left of $\pi$ and $< 0$ just right. Hence $\cot x \to \frac{-1}{0^+} = -\infty$ as $x \to \pi^-$ and $\frac{-1}{0^-} = +\infty$ as $x \to \pi^+$ : $x = \pi$ is a vertical asymptote. The sign of $\sin$ near $\pi$ comes from the unit circle, not from the calculator.

Exercise 4: Exponentials and logarithms at the ends of the axis

The facts read on the graphs: $e^x \to \infty$ as $x \to \infty$ and $e^x \to 0$ as $x \to -\infty$ , so $e^{-x}$ does the opposite; $\ln x \to \infty$ as $x \to \infty$ and $\ln x \to -\infty$ as $x \to 0^+$ . The algebra that goes with them: $e^{2x} = \left(e^x\right)^2$ , $\frac{e^a}{e^b} = e^{a - b}$ , $e^{-x} = \frac{1}{e^x}$ , $\ln A - \ln B = \ln\frac{A}{B}$ and $k\ln x = \ln x^k$ for $x > 0$ .

Two tools of Thomas 2.6 also appear: the Sandwich Theorem, which works as $x \to \pm\infty$ exactly as at a point, and the substitution $t = \frac{1}{x}$ , which turns $x \to \infty$ into $t \to 0^+$ . The figure shows $h(x) = \frac{5 - e^x}{1 + e^x}$ .

a) Find $\lim_{x\to\infty} h(x)$ and $\lim_{x\to -\infty} h(x)$ . Find exactly where the graph of $h$ crosses the $x$ -axis.
b) Find $\lim_{x\to\infty} \frac{e^{2x} + 1}{e^{2x} - e^x}$ and $\lim_{x\to -\infty} \frac{e^{2x} + 1}{e^{2x} - e^x}$ .
c) Using $t = \frac{1}{x}$ , find $\lim_{x\to\infty} e^{1/x}$ and $\lim_{x\to\infty} 3x\tan\left(\frac{1}{x}\right)$ .
d) Find $\lim_{x\to\infty} \left[2\ln x - \ln(4x^2 + 1)\right]$ and $\lim_{x\to 0^+} \left[2\ln x - \ln(4x^2 + 1)\right]$ .
e) Find $\lim_{x\to\infty} e^{-x}\cos x$ . What happens to $e^{-x}\cos x$ as $x \to -\infty$ ?

Type your answers, the page tells you right or wrong 0/11

a)

\lim_{x\to\infty} h(x)

\lim_{x\to -\infty} h(x)

Crossing:

x =

b)

\lim_{x\to\infty}

\lim_{x\to -\infty}

c)

\lim_{x\to\infty} e^{1/x}

\lim_{x\to\infty} 3x\tan\frac{1}{x}

d)

\lim_{x\to\infty}

\lim_{x\to 0^+}

e)

\lim_{x\to\infty} e^{-x}\cos x

\lim_{x\to -\infty} e^{-x}\cos x

Show the solution

Answers

a) $-1$ and $5$ ; the graph crosses the $x$ -axis at $x = \ln 5 \approx 1.6094$ .
b) $1$ and $-\infty$
c) $1$ and $3$
d) $\ln\frac{1}{4} = -\ln 4 \approx -1.3863$ , and $-\infty$ (not an indeterminate form).
e) $0$ by the Sandwich Theorem; as $x \to -\infty$ it oscillates with growing amplitude: no limit, not even an infinite one.

a) As $x \to \infty$ , $e^x$ dominates: divide every term by it. $h(x) = \frac{5e^{-x} - 1}{e^{-x} + 1} \to \frac{0 - 1}{0 + 1} = -1$ . As $x \to -\infty$ , $e^x \to 0$ and nothing needs dividing: $h(x) \to \frac{5 - 0}{1 + 0} = 5$ . Two horizontal asymptotes, $y = 5$ on the left and $y = -1$ on the right. The crossing: $h(x) = 0$ exactly when $5 - e^x = 0$ , since $1 + e^x > 0$ ; so $e^x = 5$ and $x = \ln 5 \approx 1.6094$ . Note that $\frac{5e^{-x} - 1}{e^{-x} + 1}$ comes from $\frac{5}{e^x} = 5e^{-x}$ and $\frac{e^x}{e^x} = 1$ : each term divided, each exponent subtracted.

b) As $x \to \infty$ the dominant exponential is $e^{2x}$ : $\frac{1 + e^{-2x}}{1 - e^{-x}} \to 1$ , because $\frac{e^x}{e^{2x}} = e^{x - 2x} = e^{-x}$ . As $x \to -\infty$ , the numerator tends to $0 + 1 = 1$ and the denominator to $0 - 0 = 0$ : study its sign. Factor: $e^{2x} - e^x = e^x\left(e^x - 1\right)$ , with $e^x > 0$ and $e^x - 1 < 0$ for $x < 0$ , so the denominator tends to $0^-$ and the quotient to $-\infty$ . Dividing by $e^{2x}$ at $-\infty$ would be dividing by something that tends to $0$ : the dominant term depends on the direction.

c) With $t = \frac{1}{x}$ , $x \to \infty$ means $t \to 0^+$ . First, $e^{1/x} = e^t \to e^0 = 1$ , by continuity of the exponential. Second, $3x\tan\left(\frac{1}{x}\right) = \frac{3\tan t}{t} = 3 \cdot \frac{\sin t}{t} \cdot \frac{1}{\cos t} \to 3 \cdot 1 \cdot 1 = 3$ , using $\lim_{t\to 0} \frac{\sin t}{t} = 1$ from section 2.4. As written, the limit is the form $\infty \cdot 0$ ( $x \to \infty$ , $\tan\frac{1}{x} \to 0$ ), which decides nothing; the substitution turns it into a limit we know.

d) As $x \to \infty$ both logarithms tend to $\infty$ : the form $\infty - \infty$ . Use the laws of logarithms, valid for $x > 0$ : $2\ln x = \ln x^2$ and $\ln x^2 - \ln(4x^2 + 1) = \ln\frac{x^2}{4x^2 + 1}$ . The inside tends to $\frac{1}{4}$ (equal degrees), and $\ln$ is continuous at $\frac{1}{4}$ , so the limit is $\ln\frac{1}{4} = -\ln 4 \approx -1.3863$ . The false law $\ln(4x^2 + 1) = \ln 4x^2 + \ln 1$ is the slip to avoid: the logarithm of a SUM does not split. As $x \to 0^+$ , $2\ln x \to -\infty$ while $\ln(4x^2 + 1) \to \ln 1 = 0$ : the difference tends to $-\infty$ , and there is nothing indeterminate to transform.

e) For every $x$ , $-1 \le \cos x \le 1$ , and $e^{-x} > 0$ , so $-e^{-x} \le e^{-x}\cos x \le e^{-x}$ . Both bounds tend to $0$ as $x \to \infty$ , so by the Sandwich Theorem $\lim_{x\to\infty} e^{-x}\cos x = 0$ : $y = 0$ is a horizontal asymptote, crossed at every zero of $\cos x$ . As $x \to -\infty$ , $e^{-x} \to \infty$ : at $x = -2k\pi$ the function equals $e^{2k\pi}$ , which grows without bound, and at $x = -(2k+1)\pi$ it equals $-e^{(2k+1)\pi}$ , which goes to $-\infty$ . The values are neither close to one number nor all large and positive: the limit does not exist, not even as $\infty$ .

Exercise 5: Oblique asymptotes and dominant terms: long division done right

When the degree of the numerator of a rational function is exactly ONE more than the degree of the denominator, the graph has an oblique (slant) asymptote. Long division gives $f(x) = (mx + b) + \frac{R(x)}{D(x)}$ with $\deg R < \deg D$ ; the fraction tends to $0$ at both ends, and Thomas calls $mx + b$ the dominant term of $f$ for large $|x|$ . Near a vertical asymptote, the dominant term is the one that blows up.

Two algebra rules make or break the division: a missing power is written with a $0$ placeholder ( $4x^3 + 0x^2 - 2x + 5$ ), and subtracting a line changes the sign of EVERY term of that line. The figure shows $f(x) = \frac{2x^2 - 3x - 1}{x - 2}$ and the dashed line $y = 2x + 1$ .

a) Divide $2x^2 - 3x - 1$ by $x - 2$ and deduce the oblique asymptote of $f$ . Give its vertical asymptote with the two one-sided limits.
b) On which side of its oblique asymptote is the graph of $f$ , for $x > 2$ and for $x < 2$ ?
c) Let $q(x) = \frac{4x^3 - 2x + 5}{2x^2 - x}$ . Find its oblique asymptote and the point where the graph crosses it. How many vertical asymptotes does $q$ have?
d) Let $s(x) = \frac{x^3 - 4x + 6}{x^2}$ . Write $s(x)$ as a sum of terms, then give its dominant term for large $|x|$ and its behaviour near $x = 0$ .
e) Let $u(x) = 2x - 3 + e^{-x}$ . Is the line $y = 2x - 3$ an asymptote of $u$ as $x \to \infty$ ? As $x \to -\infty$ ?

Type your answers, the page tells you right or wrong 0/18

a)

Oblique asymptote: slope

Oblique asymptote: intercept

Vertical asymptote:

x =

\lim_{x\to 2^-} f(x)

\lim_{x\to 2^+} f(x)

b)

For

x > 2

the graph is

For

x < 2

the graph is

c)

Oblique asymptote: slope

Oblique asymptote: intercept

Crossing point:

x

Crossing point:

y

Number of vertical asymptotes

d)

Oblique asymptote: slope

Oblique asymptote: intercept

\lim_{x\to 0^-} s(x)

\lim_{x\to 0^+} s(x)

e)

Asymptote as

x \to \infty

?

Asymptote as

x \to -\infty

?

Show the solution

Answers

a) $f(x) = 2x + 1 + \frac{1}{x - 2}$ : oblique asymptote $y = 2x + 1$ ; vertical asymptote $x = 2$ , $-\infty$ on the left, $+\infty$ on the right.
b) Above for $x > 2$ , below for $x < 2$ , since $f(x) - (2x + 1) = \frac{1}{x - 2}$ .
c) $q(x) = 2x + 1 + \frac{5 - x}{2x^2 - x}$ : asymptote $y = 2x + 1$ , crossed at $(5, 11)$ ; two vertical asymptotes, $x = 0$ and $x = \frac{1}{2}$ .
d) $s(x) = x - \frac{4}{x} + \frac{6}{x^2}$ : dominant term $x$ (asymptote $y = x$ ); $s(x) \to +\infty$ on both sides of $0$ .
e) Yes as $x \to \infty$ (the gap $e^{-x} \to 0$ ); no as $x \to -\infty$ (the gap $e^{-x} \to \infty$ ).

a) Long division: $2x^2 \div x = 2x$ , and $2x^2 - 3x - 1 - 2x(x - 2) = 2x^2 - 3x - 1 - 2x^2 + 4x = x - 1$ ; then $x \div x = 1$ , and $x - 1 - 1 \cdot (x - 2) = 1$ . So $2x^2 - 3x - 1 = (x - 2)(2x + 1) + 1$ and $f(x) = 2x + 1 + \frac{1}{x - 2}$ . Check at $x = 3$ : $f(3) = \frac{18 - 9 - 1}{1} = 8$ and $6 + 1 + 1 = 8$ . As $x \to \pm\infty$ , $\frac{1}{x - 2} \to 0$ , so $y = 2x + 1$ is the oblique asymptote at both ends. At $x = 2$ the numerator is $8 - 6 - 1 = 1 \ne 0$ : with $x - 2 \to 0^-$ on the left, $f \to -\infty$ ; with $0^+$ on the right, $f \to +\infty$ . The subtraction step is where the division goes wrong: $-2x(x - 2) = -2x^2 + 4x$ , and forgetting to change the sign of $-4x$ gives the wrong asymptote $y = 2x - 7$ .

b) $f(x) - (2x + 1) = \frac{1}{x - 2}$ , positive for $x > 2$ and negative for $x < 2$ . So the right branch lies ABOVE the dashed line and comes down onto it, and the left branch lies BELOW it and climbs to it, exactly as on the figure. The sign of the remainder term is what makes a sketch near an oblique asymptote correct.

c) Write the missing power: $4x^3 + 0x^2 - 2x + 5$ . First step: $4x^3 \div 2x^2 = 2x$ , and $2x(2x^2 - x) = 4x^3 - 2x^2$ ; subtracting leaves $0x^2 + 2x^2 - 2x + 5 = 2x^2 - 2x + 5$ . Second step: $2x^2 \div 2x^2 = 1$ , and subtracting $2x^2 - x$ leaves $-x + 5$ . So $q(x) = 2x + 1 + \frac{5 - x}{2x^2 - x}$ , and since the fraction tends to $0$ (degree $1$ over $2$ ), $y = 2x + 1$ is the oblique asymptote. The gap vanishes when $5 - x = 0$ : at $x = 5$ , $q(5) = \frac{500 - 10 + 5}{50 - 5} = \frac{495}{45} = 11 = 2 \cdot 5 + 1$ , so the graph crosses its asymptote at $(5, 11)$ . The denominator $x(2x - 1)$ vanishes at $0$ and $\frac{1}{2}$ , where the numerator equals $5$ and $\frac{1}{2} - 1 + 5 = 4.5$ , both nonzero: two vertical asymptotes. Without the $0x^2$ placeholder, the $-2x^2$ of the first subtraction lands under $-2x$ and the division collapses.

d) The denominator is a single term, so divide EACH term of the numerator by it: $s(x) = \frac{x^3}{x^2} - \frac{4x}{x^2} + \frac{6}{x^2} = x - \frac{4}{x} + \frac{6}{x^2}$ . For large $|x|$ the last two terms tend to $0$ : the dominant term is $x$ , and $y = x$ is the oblique asymptote at both ends. Near $x = 0$ the dominant term is $\frac{6}{x^2}$ , positive on both sides, so $s(x) \to +\infty$ as $x \to 0^-$ and as $x \to 0^+$ (directly: the numerator tends to $6$ and $x^2 \to 0^+$ ). The slip to avoid is dividing only the first term: $\frac{x^3 - 4x + 6}{x^2} = x - 4x + 6$ is false.

e) $u(x) - (2x - 3) = e^{-x}$ . As $x \to \infty$ , $e^{-x} \to 0$ , so $y = 2x - 3$ is an oblique asymptote on the right, approached from above since $e^{-x} > 0$ . As $x \to -\infty$ , $e^{-x} \to \infty$ : the gap grows without bound, so the line is NOT an asymptote on the left. An oblique asymptote, like a horizontal one, belongs to ONE direction at a time; rational functions hide this because their two ends always agree.

Exercise 6: Asymptotes as equations: finding the constants

A final-exam favourite turns the chapter around: the asymptotes are given, the formula carries unknown constants, and each asymptote becomes an equation. Every candidate vertical asymptote must then be checked in the numerator, since a common zero gives a hole instead.

No figure: every answer comes from the formula.

a) Find $a$ and $b$ so that the graph of $f(x) = \frac{ax^2 - 3}{x^2 + bx}$ has the horizontal asymptote $y = -2$ and the vertical asymptote $x = 4$ . Does $f$ have another vertical asymptote?
b) Let $h(x) = \frac{x - k}{x^2 - 5x + 6}$ . For which values of $k$ does the graph of $h$ have exactly ONE vertical asymptote? For the smaller such $k$ , describe the graph near its other excluded point.
c) Find $a$ and $b > 0$ so that the graph of $g(x) = \frac{ae^x + 6}{e^x + b}$ has the horizontal asymptote $y = 4$ as $x \to \infty$ and $y = 3$ as $x \to -\infty$ . How many vertical asymptotes does $g$ have?
d) Find $k$ so that the graph of $p(x) = \frac{kx + 1}{\sqrt{4x^2 + 3}}$ has the horizontal asymptote $y = 5$ as $x \to -\infty$ . What is then its horizontal asymptote as $x \to \infty$ ?

Show the solution

Answers

a) $a = -2$ , $b = -4$ : $f(x) = \frac{-2x^2 - 3}{x(x - 4)}$ , with a second vertical asymptote $x = 0$ .
b) $k = 2$ or $k = 3$ . For $k = 2$ : $h(x) = \frac{1}{x - 3}$ for $x \ne 2$ , a hole at $(2, -1)$ and the asymptote $x = 3$ .
c) $a = 4$ , $b = 2$ ; no vertical asymptote, since $e^x + 2 > 0$ .
d) $k = -10$ ; then $y = -5$ as $x \to \infty$ .

a) Equal degrees, so $f(x) \to \frac{a}{1} = a$ at both ends: $a = -2$ . A vertical asymptote at $4$ needs the denominator to vanish there: $16 + 4b = 0$ , so $b = -4$ and the denominator is $x^2 - 4x = x(x - 4)$ . The numerator $-2x^2 - 3$ is negative for every $x$ , so it vanishes neither at $4$ nor at $0$ : both $x = 4$ and $x = 0$ are vertical asymptotes, and there is no hole. The factorization of the denominator is what reveals the second asymptote, which the question does not announce.

b) Factor: $x^2 - 5x + 6 = (x - 2)(x - 3)$ , so the candidates are $x = 2$ and $x = 3$ . For a generic $k$ the numerator $x - k$ vanishes at neither, and there are two vertical asymptotes. If $k = 2$ , the factor cancels: $h(x) = \frac{x - 2}{(x - 2)(x - 3)} = \frac{1}{x - 3}$ for $x \ne 2$ , which has only the asymptote $x = 3$ , and at $x = 2$ a hole at $\left(2, \frac{1}{2 - 3}\right) = (2, -1)$ . If $k = 3$ , likewise $h(x) = \frac{1}{x - 2}$ for $x \ne 3$ , a hole at $(3, 1)$ and the asymptote $x = 2$ . So exactly one vertical asymptote for $k = 2$ and for $k = 3$ , and for no other $k$ .

c) As $x \to \infty$ , divide every term by $e^x$ : $\frac{a + 6e^{-x}}{1 + be^{-x}} \to a$ , so $a = 4$ . As $x \to -\infty$ , $e^x \to 0$ : $g(x) \to \frac{0 + 6}{0 + b} = \frac{6}{b}$ , so $\frac{6}{b} = 3$ and $b = 2$ . Then $e^x + 2 > 2$ never vanishes, so $g$ has no vertical asymptote. Had the conditions forced $b = -2$ , the denominator would vanish at $x = \ln 2$ and create one: the sign of $b$ matters.

d) As $x \to -\infty$ , $\sqrt{4x^2 + 3} = |x|\sqrt{4 + \frac{3}{x^2}} = -x\sqrt{4 + \frac{3}{x^2}}$ , so $p(x) = \frac{x\left(k + \frac{1}{x}\right)}{-x\sqrt{4 + \frac{3}{x^2}}} = -\frac{k + \frac{1}{x}}{\sqrt{4 + \frac{3}{x^2}}} \to -\frac{k}{2}$ . Setting $-\frac{k}{2} = 5$ gives $k = -10$ . As $x \to \infty$ the minus sign disappears and $p(x) \to \frac{k}{2} = -5$ . A student who forgets $|x| = -x$ finds $k = 10$ and swaps the two asymptotes: check with $x = -100$ , where $p(-100) = \frac{1001}{\sqrt{40003}} > 0$ , consistent with $y = 5$ on the left.

Exercise 7: Creating graphs from limits, and limits from graphs

The figure shows the whole graph of a function $f$ defined on $(-\infty, -1) \cup (-1, \infty)$ : dashed lines mark $x = -1$ , $x = 3$ , $y = 2$ and $y = -1$ , and the dot is the point of the graph at $x = 3$ .

Thomas's exercises on creating graphs work in both directions: read the limits on a graph, and choose a formula that produces given limits. A formula is checked limit by limit, never at a glance.

a) Read on the graph $\lim_{x\to -\infty} f(x)$ , $\lim_{x\to\infty} f(x)$ , $\lim_{x\to -1^-} f(x)$ , $\lim_{x\to -1^+} f(x)$ and $\lim_{x\to 3} f(x)$ .
b) Give $f(3)$ . Is $x = 3$ a vertical asymptote, although $f$ is defined there? How many asymptotes does the graph have in all?
c) Which function satisfies ALL of: the only vertical asymptote is $x = 1$ , with $\lim_{x\to 1^\pm} = +\infty$ ; $\lim_{x\to\pm\infty} = 3$ ; and the value $0$ at $x = 0$ ? Choose among $\frac{3x^2}{(x-1)^2}$ , $\frac{3x}{(x-1)^2}$ , $\frac{3x^2}{x^2 - 1}$ , $\frac{3(x-1)^2}{x^2}$ and $\frac{3x^2 + 1}{(x-1)^2}$ .
d) Which function has $\lim_{x\to\pm\infty} = 0$ , $\lim_{x\to 2^-} = -\infty$ , $\lim_{x\to 2^+} = +\infty$ , $\lim_{x\to -2^-} = +\infty$ and $\lim_{x\to -2^+} = -\infty$ ? Choose among $\frac{1}{x^2 - 4}$ , $\frac{x}{x^2 - 4}$ , $\frac{x^2}{x^2 - 4}$ and $\frac{1}{(x - 2)^2(x + 2)}$ .
e) Could the function of the figure be a rational function (a quotient of two polynomials)? Explain.

Type your answers, the page tells you right or wrong 0/11

a)

\lim_{x\to -\infty} f(x)

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

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

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

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

b)

f(3)

Is

x = 3

a vertical asymptote?

Number of asymptotes

c)

The function

d)

The function

e)

Could

f

be a rational function?

Show the solution

Answers

a) $2$ ; $-1$ ; $+\infty$ ; $-\infty$ ; $+\infty$ (from both sides).
b) $f(3) = 1$ ; yes, $x = 3$ is a vertical asymptote; four asymptotes: $x = -1$ , $x = 3$ , $y = 2$ , $y = -1$ .
c) $\frac{3x^2}{(x-1)^2}$
d) $\frac{1}{x^2 - 4}$
e) No: a rational function has the same limit at $\infty$ and at $-\infty$ (or none), and this graph has two different horizontal asymptotes.

a) Far to the left the curve settles on the dashed line $y = 2$ , so $\lim_{x\to -\infty} f(x) = 2$ ; far to the right it settles on $y = -1$ , so $\lim_{x\to\infty} f(x) = -1$ . Along $x = -1$ the left branch climbs out of the frame and the middle branch comes up from the bottom: $\lim_{x\to -1^-} f(x) = +\infty$ and $\lim_{x\to -1^+} f(x) = -\infty$ . Along $x = 3$ both neighbouring branches climb out of the frame: $\lim_{x\to 3^-} f(x) = \lim_{x\to 3^+} f(x) = +\infty$ , so $\lim_{x\to 3} f(x) = +\infty$ .

b) The dot gives $f(3) = 1$ . The line $x = 3$ is nevertheless a vertical asymptote: the definition asks about the one-sided LIMITS at $3$ , which are infinite, and never about the value at $3$ . The graph has four asymptotes: the vertical lines $x = -1$ and $x = 3$ , and the horizontal lines $y = 2$ (on the left) and $y = -1$ (on the right).

c) Test each condition in turn. $\frac{3x^2}{(x-1)^2}$ : equal degrees with leading coefficients $3$ and $1$ , so the limit is $3$ at both ends; near $1$ the numerator tends to $3 > 0$ and $(x-1)^2 \to 0^+$ on both sides, so $+\infty$ on both sides; the only zero of the denominator is $1$ ; and its value at $0$ is $0$ . It satisfies everything. The others fail: $\frac{3x}{(x-1)^2}$ tends to $0$ at infinity (degree $1$ over $2$ ); $\frac{3x^2}{x^2 - 1}$ has a second asymptote $x = -1$ and changes sign at $1$ ; $\frac{3(x-1)^2}{x^2}$ has its asymptote at $x = 0$ , where it is not even defined; $\frac{3x^2 + 1}{(x-1)^2}$ equals $1$ at $x = 0$ .

d) $\frac{1}{x^2 - 4} = \frac{1}{(x - 2)(x + 2)}$ tends to $0$ at both ends. Near $2$ , $x + 2 \to 4$ and $x - 2 \to 0^\mp$ : $-\infty$ on the left, $+\infty$ on the right. Near $-2$ , $x - 2 \to -4$ and $x + 2 \to 0^-$ on the left, so the product tends to $0^+$ and the quotient to $+\infty$ ; on the right the product tends to $0^-$ and the quotient to $-\infty$ . All five conditions hold. $\frac{x}{x^2 - 4}$ fails at $-2$ : its numerator tends to $-2 < 0$ , which reverses both signs there. $\frac{x^2}{x^2 - 4}$ tends to $1$ at infinity. $\frac{1}{(x-2)^2(x+2)}$ tends to $+\infty$ on BOTH sides of $2$ , because of the square.

e) No. For a rational function $\frac{P}{Q}$ , dividing by the highest power of the denominator gives the same result as $x \to \infty$ and as $x \to -\infty$ when the limit is finite: $0$ , or the ratio of the leading coefficients. So a rational function has at most ONE horizontal asymptote, shared by both ends. The graph of the figure approaches $y = 2$ on the left and $y = -1$ on the right, so it is not the graph of a rational function: a root, an exponential or a piecewise definition is needed (here the three branches come from three different formulas).

Exercise 8: Five statements to correct

Each statement below was written on a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample or the correct computation, and write the correct statement. Four of the five errors are algebra, not calculus.

a) $\sqrt{x^2 + 5} = x + \sqrt{5}$ for $x > 0$ , so $\lim_{x\to\infty} \left(\sqrt{x^2 + 5} - x\right) = \sqrt{5}$ .
b) The graph of $\frac{x - 3}{x^2 - 9}$ has two vertical asymptotes, $x = 3$ and $x = -3$ .
c) $\lim_{x\to\infty} \frac{e^{2x}}{e^x + 1} = e^2$ , because $\frac{e^{2x}}{e^x} = e^2$ .
d) If $\lim_{x\to\infty} f(x) = 0$ and $\lim_{x\to\infty} g(x) = \infty$ , then $\lim_{x\to\infty} f(x)g(x) = 0$ .
e) The graph of $y = \ln x$ flattens out, so it has a horizontal asymptote as $x \to \infty$ .

Type your answers, the page tells you right or wrong 0/9

a)

The statement is

The correct

\lim_{x\to\infty} (\sqrt{x^2 + 5} - x)

b)

The statement is

Number of vertical asymptotes

c)

The statement is

The correct limit

d)

The statement is

e)

The statement is

\lim_{x\to\infty} \ln x

Show the solution

Answers

a) False: $\sqrt{a + b} \ne \sqrt a + \sqrt b$ ; by the conjugate, $\sqrt{x^2 + 5} - x = \frac{5}{\sqrt{x^2 + 5} + x} \to 0$ .
b) False: $\frac{x - 3}{x^2 - 9} = \frac{1}{x + 3}$ for $x \ne 3$ ; a hole at $\left(3, \frac{1}{6}\right)$ and only one asymptote, $x = -3$ .
c) False: $\frac{e^{2x}}{e^x} = e^x$ ; the quotient equals $\frac{e^x}{1 + e^{-x}} \to \infty$ .
d) False: $0 \cdot \infty$ is indeterminate; $\frac{1}{x} \cdot x^2 \to \infty$ , $\frac{3}{x} \cdot x \to 3$ .
e) False: $\ln x > M$ as soon as $x > e^M$ , so $\ln x \to \infty$ : no horizontal asymptote.

a) FALSE. A square root does not split over a sum: at $x = 2$ , $\sqrt{4 + 5} = 3$ while $2 + \sqrt 5 \approx 4.24$ . The limit is a form $\infty - \infty$ to settle by the conjugate: $\sqrt{x^2 + 5} - x = \frac{(x^2 + 5) - x^2}{\sqrt{x^2 + 5} + x} = \frac{5}{\sqrt{x^2 + 5} + x}$ , whose denominator tends to $\infty$ , so the limit is $0$ . Correct statement: $\sqrt{x^2 + 5} - x \to 0$ as $x \to \infty$ ; in general $\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$ .

b) FALSE. Factor before deciding: $x^2 - 9 = (x - 3)(x + 3)$ , so $\frac{x - 3}{x^2 - 9} = \frac{1}{x + 3}$ for $x \ne 3$ and the limit at $3$ is $\frac{1}{6}$ : the graph has a hole at $\left(3, \frac{1}{6}\right)$ , not an asymptote. At $-3$ the simplified numerator is $1 \ne 0$ , so $x = -3$ is the only vertical asymptote. Correct statement: a zero of the denominator gives a vertical asymptote only if the numerator does not tend to $0$ there.

c) FALSE. The exponent law is $\frac{e^a}{e^b} = e^{a - b}$ , so $\frac{e^{2x}}{e^x} = e^{2x - x} = e^x$ , not $e^2$ (that would be dividing the exponents and dropping $x$ ). Dividing every term by $e^x$ : $\frac{e^{2x}}{e^x + 1} = \frac{e^x}{1 + e^{-x}}$ , whose numerator tends to $\infty$ and denominator to $1$ . Correct statement: $\lim_{x\to\infty} \frac{e^{2x}}{e^x + 1} = \infty$ , no horizontal asymptote on the right.

d) FALSE. $0 \cdot \infty$ is an indeterminate form: the answer depends on how fast each factor goes. With $f(x) = \frac{1}{x}$ and $g(x) = x^2$ , $fg = x \to \infty$ ; with $f(x) = \frac{3}{x}$ and $g(x) = x$ , $fg = 3$ ; with $f(x) = \frac{1}{x^2}$ and $g(x) = x$ , $fg = \frac{1}{x} \to 0$ . Correct statement: the product must be simplified into one expression before its limit is taken.

e) FALSE. $\ln x$ grows slowly but WITHOUT BOUND: given any level $M$ , $\ln x > M$ for every $x > e^M$ , since $\ln$ is increasing. For instance $\ln x > 100$ for $x > e^{100}$ . So $\lim_{x\to\infty} \ln x = \infty$ and no horizontal line is approached. The flattening seen on a calculator window is an effect of the window. Correct statement: $\ln x \to \infty$ as $x \to \infty$ ; the graph of $\ln x$ has only the vertical asymptote $x = 0$ .

Exercise 9: Average cost: what happens when production grows

A workshop assembles electric bicycles. Its fixed costs are $18\,000$ dollars a month (rent, salaries, equipment), and the parts of each bicycle cost $240$ dollars. In a first model, producing $x$ bicycles in a month costs $C_1(x) = 18\,000 + 240x$ dollars. The AVERAGE COST per bicycle is $A(x) = \frac{C(x)}{x}$ .

Above a certain volume the workshop pays overtime and rents extra space; a second model adds a term for that: $C_2(x) = 18\,000 + 240x + 0.05x^2$ . A third option is an automated line, with $C_3(x) = 60\,000 + 200x + 3000\sqrt{x}$ . The figure shows the average costs $A_1$ and $A_2$ of the first two models. Give amounts to the cent when they are not whole.

a) Write $A_1(x)$ as a sum of simple terms. Find $\lim_{x\to\infty} A_1(x)$ and $\lim_{x\to 0^+} A_1(x)$ , and interpret both.
b) Show that $A_1(x) > 240$ for every $x > 0$ . What is the smallest monthly production for which the average cost is below $250$ dollars?
c) Write $A_2(x)$ as a sum of simple terms and show that its graph has an oblique asymptote. Give it.
d) Find $\lim_{x\to\infty} A_2(x)$ . For which productions is $A_2(x)$ within $5$ dollars of its oblique asymptote?
e) Find $\lim_{x\to\infty} A_3(x)$ and $A_3(500)$ . Which option has the lower average cost in the long run, the automated line or model 1?

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

a)

\lim_{x\to\infty} A_1(x)

dollars

\lim_{x\to 0^+} A_1(x)

b)

Smallest production bicycles

c)

Oblique asymptote: slope

Oblique asymptote: intercept

d)

\lim_{x\to\infty} A_2(x)

Within 5 dollars for

x >

bicycles

e)

\lim_{x\to\infty} A_3(x)

dollars

A_3(500)

dollars

Lower average cost in the long run

Show the solution

Answers

a) $A_1(x) = \frac{18\,000}{x} + 240$ ; $\lim_{x\to\infty} A_1 = 240$ dollars (the cost of the parts), $\lim_{x\to 0^+} A_1 = +\infty$ .
b) $A_1(x) - 240 = \frac{18\,000}{x} > 0$ ; below $250$ dollars from $x = 1801$ bicycles on.
c) $A_2(x) = \frac{18\,000}{x} + 240 + 0.05x$ ; oblique asymptote $y = 0.05x + 240$ .
d) $+\infty$ ; within $5$ dollars for $x > 3600$ bicycles.
e) $200$ dollars; $A_3(500) \approx 454.16$ dollars; the automated line, since $200 < 240$ .

a) The denominator $x$ is a single term, so divide EACH term of the cost by it: $A_1(x) = \frac{18\,000 + 240x}{x} = \frac{18\,000}{x} + 240$ . As $x \to \infty$ , $\frac{18\,000}{x} \to 0$ , so $A_1(x) \to 240$ : the line $y = 240$ is a horizontal asymptote. In words, the fixed costs are spread over more and more bicycles and their share per bicycle vanishes; what remains is the $240$ dollars of parts that every bicycle needs. As $x \to 0^+$ , $\frac{18\,000}{x} \to +\infty$ : the $y$ -axis is a vertical asymptote, and a workshop that builds almost nothing pays an enormous amount per bicycle. Writing $\frac{18\,000 + 240x}{x} = 18\,000 + 240$ (cancelling $x$ in one term only) is the algebra slip this part is testing.

b) $A_1(x) - 240 = \frac{18\,000}{x}$ , positive for every $x > 0$ : the average cost always stays above its asymptote and approaches it from above, never reaching it. Next, $A_1(x) < 250$ means $\frac{18\,000}{x} < 10$ , that is $x > 1800$ (multiplying by $x > 0$ keeps the direction of the inequality). At exactly $1800$ bicycles the average cost is $250$ dollars, not below, so the smallest production is $1801$ bicycles, where $A_1 = \frac{18\,000}{1801} + 240 \approx 249.99$ dollars.

c) Dividing each term by $x$ : $A_2(x) = \frac{18\,000}{x} + 240 + 0.05x$ . The gap between $A_2$ and the line $y = 0.05x + 240$ is $A_2(x) - (0.05x + 240) = \frac{18\,000}{x}$ , which tends to $0$ as $x \to \infty$ . So $y = 0.05x + 240$ is an oblique asymptote, approached from above. Here no long division is needed: when the denominator is a single power of $x$ , splitting the fraction term by term IS the division.

d) $A_2(x) \ge 240 + 0.05x$ , which tends to $\infty$ , so $\lim_{x\to\infty} A_2(x) = +\infty$ and there is no horizontal asymptote: for large volumes the overtime term $0.05x$ dominates and the average cost climbs again, as the figure shows. Within $5$ dollars of the asymptote means $\frac{18\,000}{x} < 5$ , that is $x > 3600$ bicycles. The dominant terms tell the whole story: $\frac{18\,000}{x}$ near $0$ , $0.05x$ for large $x$ .

e) Divide each term by $x$ , with $\frac{\sqrt{x}}{x} = x^{1/2 - 1} = x^{-1/2} = \frac{1}{\sqrt{x}}$ : $A_3(x) = \frac{60\,000}{x} + 200 + \frac{3000}{\sqrt{x}}$ . Both fractions tend to $0$ , so $\lim_{x\to\infty} A_3(x) = 200$ dollars. At $x = 500$ : $A_3(500) = 120 + 200 + \frac{3000}{\sqrt{500}} \approx 120 + 200 + 134.16 = 454.16$ dollars, while $A_1(500) = 36 + 240 = 276$ dollars. So the automated line is much more expensive at $500$ bicycles, but its asymptote $200$ is below the $240$ of model 1: in the long run it wins. At $10\,000$ bicycles, $A_3 = 6 + 200 + 30 = 236$ dollars against $A_1 = 241.80$ dollars. The limits compare the two options for very large volumes only; the values at a realistic $x$ are what decide a purchase.

Exercise 10: A drug in the long term: washout and steady state

The same drug can be given in two ways. After a single injection, the concentration in the blood is modelled by $C_A(t) = \frac{40t}{t^2 + 16}$ mg/L, $t$ hours after the injection. A skin patch releases the drug continuously, and the concentration is then modelled by $C_B(t) = \frac{20t}{\sqrt{t^2 + 25}}$ mg/L, $t$ hours after the patch is applied.

The questions are about the long term only: what each concentration tends to, and how long that takes. Round times to the hundredth of an hour. The figure shows both models over two days.

a) Find $\lim_{t\to\infty} C_A(t)$ and interpret it. What is the dominant term of $C_A(t)$ for large $t$ ?
b) From what time on does the concentration after the injection stay below $2$ mg/L?
c) Find $\lim_{t\to\infty} C_B(t)$ , the steady-state concentration of the patch. Show that it is never reached, and find when $C_B(t) = 16$ mg/L.
d) From what time on is $C_B(t)$ within $0.5$ mg/L of its steady state?
e) A patient receives the injection and the patch at the same moment, so that $C(t) = C_A(t) + C_B(t)$ . Find $\lim_{t\to\infty} C(t)$ and $\lim_{t\to\infty} \frac{C_A(t)}{C_B(t)}$ . Does $C(t)$ approach its limit from above or from below?

Show the solution

Answers

a) $0$ mg/L: the drug is eliminated; dominant term $\frac{40}{t}$ .
b) From $t = 10 + 2\sqrt{21} \approx 19.17$ h on (the other root, $0.83$ h, is during the rise).
c) $20$ mg/L; $20 - C_B(t) > 0$ since $t < \sqrt{t^2 + 25}$ ; $C_B = 16$ at $t = \frac{20}{3} \approx 6.67$ h.
d) From $t \approx 21.94$ h on.
e) $20$ mg/L and $0$ ; from above, since $C_A \approx \frac{40}{t}$ outweighs $20 - C_B \approx \frac{250}{t^2}$ .

a) Divide every term by $t^2$ , the dominant power of the denominator: $C_A(t) = \frac{\frac{40}{t}}{1 + \frac{16}{t^2}} \to \frac{0}{1} = 0$ . The line $C = 0$ is a horizontal asymptote: in the long run the drug is eliminated (washout). For large $t$ , $t^2 + 16$ is dominated by $t^2$ , so $C_A(t)$ behaves like $\frac{40t}{t^2} = \frac{40}{t}$ : at $t = 100$ , $C_A = \frac{4000}{10\,016} \approx 0.3994$ against $\frac{40}{100} = 0.4$ . The dominant term says HOW FAST the limit is approached, which the limit alone does not.

b) $C_A(t) < 2$ means $40t < 2(t^2 + 16)$ , since $t^2 + 16 > 0$ , that is $t^2 - 20t + 16 > 0$ . The roots of $t^2 - 20t + 16$ are $t = \frac{20 \pm \sqrt{400 - 64}}{2} = 10 \pm \sqrt{84} = 10 \pm 2\sqrt{21}$ , about $0.83$ and $19.17$ . The quadratic is positive outside its roots, so $C_A(t) < 2$ for $t < 0.83$ and for $t > 19.17$ . The first interval is the first minutes, while the concentration is still RISING past $2$ mg/L; the question asks when it STAYS below, so the answer is from $t = 10 + 2\sqrt{21} \approx 19.17$ h on. Keeping only one root, or the wrong one, is the usual loss here.

c) For $t > 0$ , $\sqrt{t^2 + 25} = t\sqrt{1 + \frac{25}{t^2}}$ , so $C_B(t) = \frac{20}{\sqrt{1 + \frac{25}{t^2}}} \to 20$ mg/L. Never reached: $t < \sqrt{t^2 + 25}$ for every $t$ , so $\frac{t}{\sqrt{t^2 + 25}} < 1$ and $C_B(t) < 20$ . Next, $C_B(t) = 16$ means $20t = 16\sqrt{t^2 + 25}$ ; both sides are positive for $t > 0$ , so squaring is safe: $400t^2 = 256(t^2 + 25)$ , $144t^2 = 6400$ , $t^2 = \frac{400}{9}$ and $t = \frac{20}{3} \approx 6.67$ h. Squaring is safe only because both sides are positive: always say so, since squaring can create solutions that the original equation does not have.

d) $C_B(t) > 19.5$ means $20t > 19.5\sqrt{t^2 + 25}$ ; squaring (both sides positive), $400t^2 > 380.25t^2 + 9506.25$ , so $19.75t^2 > 9506.25$ , $t^2 > 481.33$ and $t > 21.94$ h. A check by the dominant term: the conjugate gives $20 - C_B(t) = \frac{20\left(\sqrt{t^2 + 25} - t\right)}{\sqrt{t^2 + 25}} = \frac{500}{\sqrt{t^2 + 25}\left(\sqrt{t^2 + 25} + t\right)}$ , which behaves like $\frac{500}{2t^2} = \frac{250}{t^2}$ ; at $t = 21.94$ that is about $0.52$ , close to $0.5$ as expected.

e) The limit of a sum is the sum of the limits: $C(t) \to 0 + 20 = 20$ mg/L. For the ratio, $\frac{C_A(t)}{C_B(t)} = \frac{40t}{t^2 + 16} \cdot \frac{\sqrt{t^2 + 25}}{20t} = \frac{2\sqrt{t^2 + 25}}{t^2 + 16}$ , which behaves like $\frac{2t}{t^2} = \frac{2}{t} \to 0$ : in the long run the injection's share of the total becomes negligible. The side: $C(t) - 20 = C_A(t) - \left(20 - C_B(t)\right)$ , which behaves like $\frac{40}{t} - \frac{250}{t^2} = \frac{40t - 250}{t^2}$ , positive for large $t$ . So $C(t)$ overshoots the steady state and comes back DOWN to $20$ from above: at $t = 50$ , $C_A \approx 0.795$ and $C_B \approx 19.901$ , total $\approx 20.70$ mg/L. Comparing $\frac{1}{t}$ with $\frac{1}{t^2}$ is a comparison of powers, the only kind this chapter needs.

Corrected exercises: limits involving infinity and asymptotes (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: Limits at infinity: divide EVERY term by the dominant power

Exercise 2: Square roots at infinity: the root of x squared is |x|

Exercise 3: Infinite limits: factor the denominator, then read the sign on each side

Exercise 4: Exponentials and logarithms at the ends of the axis

Exercise 5: Oblique asymptotes and dominant terms: long division done right

Part B: problems and reasoning (/50)

Exercise 6: Asymptotes as equations: finding the constants

Exercise 7: Creating graphs from limits, and limits from graphs

Exercise 8: Five statements to correct

Exercise 9: Average cost: what happens when production grows

Exercise 10: A drug in the long term: washout and steady state

See also

Struggling with MATH 203?