AP Calculus AB Infinite Limits and Asymptotes Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Infinite Limits and Asymptotes – Maths PDF

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This asymptotes practice solutions for AP Calculus AB covers infinite limits and asymptotes (Common Core AP CALC AB 1.13, AP CALC AB 1.14, AP CALC AB 1.15), with a free PDF to print.

This answer key gives a full worked solution for every problem in the asymptotes worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 1.13, AP CALC AB 1.14, AP CALC AB 1.15

Asymptotes practice solutions: exercises 1 to 10

Exercise 1 – One-sided infinite limits

For $x > 3$ close to 3, $x-3$ is a small positive number, so $\frac{1}{x-3}\to\infty$.

For $x < 3$, $x-3$ is a small negative number, so the quotient goes to $-\infty$.

Final answer: $\infty$ from the right, $-\infty$ from the left, and the two-sided limit does not exist.

Exercise 2 – An even power

The square is positive on both sides of $2$ and approaches $0$, so the quotient is large and positive from both sides.

Final answer: $\infty$ (the outputs grow without bound, so there is a vertical asymptote at $x=2$).

Exercise 3 – Sign analysis

Near $x=-1$, the numerator $x+3$ is about $2$ (positive).

From the left, $x+1 < 0$ and small, so the quotient is negative and large: $-\infty$. From the right, $x+1 > 0$, so the quotient goes to $\infty$.

Final answer: $-\infty$ from the left and $\infty$ from the right.

Exercise 4 – Asymptote or hole?

Factor: $\frac{x+2}{(x-3)(x+2)}$. The factor $x+2$ cancels, leaving $\frac{1}{x-3}$.

At $x=3$ the denominator is zero and the numerator is $1$, so there is a vertical asymptote. At $x=-2$ there is a hole at $\left(-2,-\frac{1}{5}\right)$.

Final answer: vertical asymptote x = 3 and a hole at x = -2.

Simplified graph: asymptote at x = 3, hole at x = -2.

Exercise 5 – Asymptotes from a graph

Left of $-1$ the branch falls: $-\infty$. Right of $-1$ it rises: $\infty$.

As $|x|$ grows, $\frac{2}{x+1}\to 0$ from both directions.

Final answer: $-\infty,\ \infty,\ 0,\ 0$ (asymptotes $x=-1$ and $y=0$).

Exercise 6 – Equal degrees

Divide top and bottom by $x^{2}$: $\frac{3+\frac{2}{x}}{5-\frac{1}{x^{2}}}$. As $x\to\infty$, $\frac{2}{x}\to 0$ and $\frac{1}{x^{2}}\to 0$.

Final answer: $\frac{3}{5}$.

Exercise 7 – Difference of roots at infinity

Multiply by the conjugate: $\frac{(9x^{2}+1)-9x^{2}}{\sqrt{9x^{2}+1}+3x}=\frac{1}{\sqrt{9x^{2}+1}+3x}$.

The denominator grows without bound, so the quotient tends to $0$.

Final answer: $0$.

Exercise 8 – Larger degree on top

The numerator degree $3$ exceeds the denominator degree $2$. Dividing by $x^{2}$: $\frac{x+\frac{1}{x^{2}}}{2+\frac{1}{x}}$, whose numerator grows without bound.

Final answer: $\infty$ (no horizontal asymptote).

Exercise 9 – Toward negative infinity

Both degrees are $3$. The ratio of leading coefficients is $\frac{2}{1}$. Dividing by $x^{3}$: $\frac{2-\frac{1}{x^{2}}}{1+\frac{4}{x^{3}}}\to 2$.

Final answer: $2$.

Exercise 10 – A square root

For large $|x|$, $\sqrt{x^{2}+1}\approx |x|$. As $x\to\infty$, $|x|=x$, so the ratio approaches $\frac{x}{x}=1$.

As $x\to-\infty$, $|x|=-x$, so the ratio approaches $\frac{-x}{x}=-1$.

Final answer: $1$ as $x\to\infty$ and $-1$ as $x\to-\infty$.

Exercises 11 and up

Exercise 11 – Horizontal asymptote and a table

The degrees are equal, so the limit is the ratio of leading coefficients: $\frac{6}{2}=3$.

The table values $2.931, 2.9993, 3, 3$ climb toward $3$.

Final answer: $y=3$.

Exercise 12 – Exponential and log limits

$e^{-x}=\frac{1}{e^{x}}\to 0$. $0$
As $x$ becomes very negative, $e^{x}$ is a tiny positive number: $0$
The logarithm grows slowly but without bound: $\infty$
Inputs near zero from the right have large negative logs: $-\infty$

Exercise 13 – A constant plus a fraction

As $x\to\infty$, $\frac{2}{x}\to 0$, so the expression approaches $5-0=5$.

Final answer: $5$.

Exercise 14 – Squeeze at infinity

Since $-1\le\sin x\le 1$, for $x > 0$ we have $-\frac{1}{x}\le\frac{\sin x}{x}\le\frac{1}{x}$.

Both bounds approach $0$ as $x\to\infty$. Final answer: $0$.

Exercise 15 – Two horizontal asymptotes

For $x > 0$, $\sqrt{x^{2}+4}\approx x$, so $g(x)\to 3$. For $x < 0$, $\sqrt{x^{2}+4}\approx -x$, so $g(x)\to -3$.

Final answer: $y=3$ as $x\to\infty$ and $y=-3$ as $x\to-\infty$.

Exercise 16 – All asymptotes

Factor: $\frac{2(x-2)(x+2)}{(x-3)(x+2)}=\frac{2(x-2)}{x-3}$ for $x\neq-2$.

Vertical asymptote: $x=3$ (numerator $2(1)=2\neq 0$). Hole: at $x=-2$, where $\frac{2(-4)}{-5}=\frac{8}{5}$.

Horizontal: equal degrees, $\frac{2}{1}=2$.

Final answer: VA x = 3, HA y = 2, hole at (-2, 8/5).

All asymptotes and the hole of f.

Exercise 17 – End behavior of a polynomial

For large $|x|$, the leading term $-2x^{3}$ dominates.

As $x\to\infty$, $-2x^{3}\to-\infty$. As $x\to-\infty$, $-2x^{3}\to\infty$.

Final answer: $\lim_{x\to\infty}p(x)=-\infty$ and $\lim_{x\to-\infty}p(x)=\infty$.

Exercise 18 – Medicine concentration

The numerator has degree $1$ and the denominator degree $2$, so $\frac{5t}{t^{2}+1}\to 0$.

Final answer: $0\text{ mg/L}$. In the long run the medicine leaves the bloodstream, and the graph has the horizontal asymptote $C=0$.

Exercise 19 – Average cost per item

Divide by $n$: $A(n)=4+\frac{200}{n}$. As $n\to\infty$, $\frac{200}{n}\to 0$.

Final answer: \$4 per sticker. For huge orders the setup fee is spread thin, so the average cost approaches the \$4 cost of each sticker.

Exercise 20 – Population limit

$P(0)=\frac{1200}{1+3}=300$.

As $t\to\infty$, $e^{-0.5t}\to 0$, so $P(t)\to\frac{1200}{1+0}=1200$.

Final answer: $P(0)=300$ trout and a long-run population of $1200$ trout.

Exercise 21 – Choosing a coefficient

The degrees are equal, so the horizontal asymptote is $y=\frac{a}{3}$.

Set $\frac{a}{3}=4$: $a=12$. Final answer: $a=12$.

Exercise 22 – Sketching with asymptotes

Vertical asymptote: $x=1$. Horizontal: $\frac{-1}{1}=-1$, so $y=-1$.

The x-intercept is where $2-x=0$: $(2,0)$. The y-intercept is $f(0)=\frac{2}{-1}=-2$: $(0,-2)$.

Final answer: VA x = 1, HA y = -1, intercepts (2, 0) and (0, -2); one branch is above $y=-1$ to the left of $x=1$ and the other below it to the right.

Graph of f(x) = (2 − x)/(x − 1).

Key facts behind this asymptotes practice solutions

Key facts

A vertical asymptote at x = a occurs when a one-sided limit there is infinite.
A horizontal asymptote y = L comes from a limit at infinity equal to L.
Rational functions: smaller top degree gives 0, equal degrees give the ratio of leading coefficients.
Check the sign on each side of a vertical asymptote to decide between +infinity and -infinity.

How to use this asymptotes practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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