AP Calculus AB Evaluating Limits Algebraically Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Evaluating Limits Algebraically – Maths PDF

Cover of the book AP Calculus AB Math Worksheets with Answers (PDF)

Download as a PDFThe AP Calculus AB worksheets book with answersPrintable PDFSee the book ›

Sommaire

This limit algebra answers for AP Calculus AB covers evaluating limits algebraically (Common Core AP CALC AB 1.5, AP CALC AB 1.6, AP CALC AB 1.7, AP CALC AB 1.8), with a free PDF to print.

This answer key gives a full worked solution for every problem in the limit algebra 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 1.5, AP CALC AB 1.6, AP CALC AB 1.7, AP CALC AB 1.8

Limit algebra answers: exercises 1 to 10

Exercise 1 – Direct substitution

The polynomial is continuous, so substitute: $3(2)^{2}-5(2)+1=12-10+1=3$.

Final answer: $3$.

Exercise 2 – Using the limit laws

$3(5)-(-2)=17$
$5\cdot(-2)=-10$
$\frac{5}{-2}=-\frac{5}{2}$
$(-2)^{2}=4$

Final answer: $17,\ -10,\ -\frac{5}{2},\ 4$.

Exercise 3 – Substitution with a root

The denominator at $x=4$ is $2$, which is not zero, so substitute directly.

$\frac{3\sqrt{4}+4}{4-2}=\frac{6+4}{2}=5$.

Final answer: $5$.

Exercise 4 – Difference of squares

Substitute $x=6$: numerator $0$, denominator $0$. The form is $\frac{0}{0}$.

Factor: $\frac{\left(x – 6\right) \left(x + 6\right)}{x – 6}$ $=x + 6$ for $x\neq 6$.

Substitute into the simplified expression: $12$.

The graph shows a hole at $(6,12)$. Final answer: $12$.

Exercise 5 – Factoring a trinomial

Substitute $x=3$: numerator $0$, denominator $0$. The form is $\frac{0}{0}$.

Factor: $\frac{\left(x – 3\right) \left(x + 4\right)}{x – 3}$ $=x + 4$ for $x\neq 3$.

Substitute into the simplified expression: $7$.

Final answer: $7$.

Simplified graph: a hole at (3, 7).

Exercise 6 – Factoring with a leading coefficient

Substitute $x=3$: numerator $0$, denominator $0$. The form is $\frac{0}{0}$.

Factor: $\frac{\left(x – 3\right) \left(2 x – 1\right)}{x – 3}$ $=2 x – 1$ for $x\neq 3$.

Substitute into the simplified expression: $5$.

Final answer: $5$.

Exercise 7 – Difference of cubes

Substitute $x=3$: numerator $0$, denominator $0$. The form is $\frac{0}{0}$.

Factor: $\frac{\left(x – 3\right) \left(x^{2} + 3 x + 9\right)}{x – 3}$ $=x^{2} + 3 x + 9$ for $x\neq 3$.

Substitute into the simplified expression: $27$.

Final answer: $27$.

Exercise 8 – Repeated factor

Substitute $x=2$: numerator $0$, denominator $0$. The form is $\frac{0}{0}$.

Factor: $\frac{\left(x – 2\right)^{2}}{\left(x – 2\right) \left(x + 2\right)}$ $=\frac{x – 2}{x + 2}$ for $x\neq 2$.

Substitute into the simplified expression: $0$.

Final answer: $0$.

Exercise 9 – Conjugate, example one

Direct substitution gives $\frac{0}{0}$. Multiply by the conjugate $\frac{\sqrt{x+5}+3}{\sqrt{x+5}+3}$:

$\frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)}=\frac{x-4}{(x-4)(\sqrt{x+5}+3)}=\frac{1}{\sqrt{x+5}+3}$.

Substitute $x=4$: $\frac{1}{3+3}=\frac{1}{6}$. Final answer: $\frac{1}{6}$.

Hole at (4, 1/6).

Exercise 10 – Conjugate, example two

Substituting gives $\frac{0}{0}$. Multiply by $\frac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3}$:

$\frac{(2x+1)-9}{(x-4)(\sqrt{2x+1}+3)}=\frac{2(x-4)}{(x-4)(\sqrt{2x+1}+3)}=\frac{2}{\sqrt{2x+1}+3}$.

Substitute: $\frac{2}{3+3}=\frac{1}{3}$. Final answer: $\frac{1}{3}$.

Exercises 11 and up

Exercise 11 – Fraction with a radical

Combine the numerator: $\frac{1}{\sqrt{x}}-\frac{1}{2}=\frac{2-\sqrt{x}}{2\sqrt{x}}$.

Write $x-4=(\sqrt{x}-2)(\sqrt{x}+2)$ and note $2-\sqrt{x}=-(\sqrt{x}-2)$: the quotient simplifies to $\frac{-1}{2\sqrt{x}(\sqrt{x}+2)}$.

Substitute $x=4$: $\frac{-1}{2\cdot 2\cdot 4}=-\frac{1}{16}$.

Final answer: $-\frac{1}{16}$.

Exercise 12 – Trig limit with a multiple

Rewrite: $\frac{\sin 5x}{x}=5\cdot\frac{\sin 5x}{5x}$. As $x\to 0$, $5x\to 0$, so $\frac{\sin 5x}{5x}\to 1$.

The limit is $5\cdot 1=5$, matching the table values near $5$. Final answer: $5$.

Exercise 13 – Tangent over a multiple

Write $\tan 5x=\frac{\sin 5x}{\cos 5x}$, so the quotient is $\frac{5}{2}\cdot\frac{\sin 5x}{5x}\cdot\frac{1}{\cos 5x}$.

As $x\to 0$, the sine factor tends to $1$ and $\cos 5x\to 1$.

Final answer: $\frac{5}{2}$.

Exercise 14 – Cosine over x sine x

Write $\frac{1-\cos 3x}{x\sin x}=\frac{1-\cos 3x}{(3x)^{2}}\cdot 9\cdot\frac{x}{\sin x}$.

Using $\lim_{u\to 0}\frac{1-\cos u}{u^{2}}=\frac{1}{2}$ (from $1-\cos u=2\sin^{2}\frac{u}{2}$) and $\frac{x}{\sin x}\to 1$, the limit is $\frac{1}{2}\cdot 9\cdot 1=\frac{9}{2}$.

Final answer: $\frac{9}{2}$.

Exercise 15 – Squeeze Theorem, cosine

For every $x\neq 0$, $-1\le\cos\frac{1}{x}\le 1$. Multiply by $x^{2}\ge 0$: $-x^{2}\le x^{2}\cos\frac{1}{x}\le x^{2}$.

Both $-x^{2}$ and $x^{2}$ approach $0$ as $x\to 0$. By the Squeeze Theorem, the limit is $0$.

The graph of sin 3x / sin 7x approaches 3/7.

Exercise 16 – Squeeze Theorem, given bounds

$\lim_{x\to 0}(3-x^{2})=3$ and $\lim_{x\to 0}(3+x^{2})=3$. Since $g$ is squeezed between them, $\lim_{x\to 0} g(x)=3$.

Exercise 17 – Finding a constant

The denominator approaches $0$, so the numerator must also approach $0$: $9+3k+6=0$, so $k=-5$.

Then $\frac{x^{2}-5x+6}{x-3}=\frac{(x-2)(x-3)}{x-3}=x-2$, and the limit is $3-2=1$.

Final answer: $k=-5$, limit $1$.

Exercise 18 – Bakery average cost

Factor: $\frac{(n-8)(n+8)}{n-8}=n+8$ for $n\neq 8$. Substitute $n=8$: $16$.

Final answer: the limit is $\$16$: the average cost approaches \$16 per cupcake near 8 dozen, even though the formula itself is undefined at 8.

Exercise 19 – Average speed over a short time

$s(2)=10$ and $s(2+h)=(2+h)^{2}+3(2+h)=4+4h+h^{2}+6+3h=10+7h+h^{2}$.

$\frac{7h+h^{2}}{h}=7+h$, which approaches $7$ as $h\to 0$.

Final answer: $7\text{ ft/s}$.

Exercise 20 – Difference quotient

$f(3)=6$. $f(3+h)=(3+h)^{2}-(3+h)=9+6h+h^{2}-3-h=6+5h+h^{2}$.

$\frac{5h+h^{2}}{h}=5+h$, so the limit as $h\to 0$ is $5$.

Exercise 21 – Water filter efficiency

Multiply by $\frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}$: $\frac{(x+16)-16}{x(\sqrt{x+16}+4)}=\frac{1}{\sqrt{x+16}+4}$.

Substitute $x=0$: $\frac{1}{4+4}=\frac{1}{8}$. Final answer: $\frac{1}{8}$.

Exercise 22 – Finding the error

The special limit needs the same expression inside the sine and in the denominator: $\frac{\sin u}{u}\to 1$. Here the sine has $2x$ but the denominator has $x$.

Rewrite $\frac{\sin 2x}{x}=2\cdot\frac{\sin 2x}{2x}$, which approaches $2\cdot 1=2$.

Final answer: Liam’s result is wrong; the limit is $2$.

Key facts behind this limit algebra answers

Key facts

Direct substitution works when the function is defined and continuous at the target.
The form 0/0 means “simplify more”, not “the limit is 0” and not “does not exist”.
Multiply by the conjugate to remove a square root from a limit.
lim sin(x)/x = 1 and lim (1 – cos x)/x = 0 as x approaches 0 (radians).

How to use this limit algebra answers

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.

Download the limit algebra answers as a PDF with the button below and keep it with your notes.

All AP Calculus AB chapters: AP Calculus AB math.

Voter.. post

Free printable PDF