# Practice Problems

 Site: Saylor Academy Course: MA005: Calculus I Book: Practice Problems
 Printed by: Guest user Date: Tuesday, July 23, 2024, 5:27 AM

## Description

Work through the odd-numbered problems 1-23. Once you have completed the problem set, check your answers.

## Problems

In problems 1–3, state each answer in the form "If $\mathrm{x}$ is within ________ units of..."

1. $\lim\limits_{x \rightarrow 3} 2 x+1=7$. What values of $x$ guarantee that $f(x)=2 x+1$ is (a) within 1 unit of 7?

(b) within $\mathrm{0.6}$ units of $\mathrm{7}$?

(c) within $\mathrm{0.4}$ units of $\mathrm{7}$?

(d) within $\varepsilon$ units of $\mathrm{7}$?

3. $\lim\limits_{x \rightarrow 2} 4 x-3=5.$ What values of $x$ guarantee that $f(x)=4 x-3$ is within $\mathrm{1}$ unit of $\mathrm{5}$?

(b) within $\mathrm{0.4}$ units of $\mathrm{5}$?

(c) within $\mathrm{0.08}$ units of $\mathrm{5}$?

(d) within $\varepsilon$ units of $\mathrm{5}$?

5. For problems 1-3, list the slope of each function $\mathrm{f}$ and the $\delta$ (as a function of $\varepsilon$). For these linear functions $\mathrm{f}$, how is $\delta$ related to the slope?

7. You have been asked to cut three boards (exactly the same length after the cut) and place them end to end. If the combined length must be within 0.06 inches of 30 inches, then each board must be within how many inches of 10?

9. $\lim\limits_{x \rightarrow 2} x^{3}=8$. What values of $x$ guarantee that $f(x)=x^{3}$ is within $0.5$ unit of 8? within $0.05$ units?

11. $\lim\limits_{\mathrm{x} \rightarrow 3} \sqrt{1+x}=2$. What values of $\mathrm{x}$ guarantee that $\mathrm{f}(\mathrm{x})=\sqrt{1+\mathrm{x}}$ is within 1 unit of 2? Within $0.0002$ units?

13. You have been asked to cut four pieces of wire (exactly the same length after the cut) and form them into a square.

If the area of the square must be within 0.06 inches of 25 inches, then each piece of wire must be within how many inches of 5?

In problems $15-17, \quad \lim\limits_{\mathrm{x} \varnothing \mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{L}$ and the function $\mathrm{f}$ and $\mathrm{a}$ value for $\varepsilon$ are given graphically. Find a length for $\delta$ that satisfies the definition of limit for the given function and value of $\varepsilon$.

15. $\mathrm{f}$ and $\varepsilon$ as shown in Fig. 17

17. $\mathrm{f}$ and $\varepsilon$ as shown in Fig. 19

In problems 19–21, use the definition to prove that the given limit does not exist.

(Find $\mathrm{a}$ value for $\varepsilon > 0$ for which there is no $\delta$ that satisfies the definition).

19. $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}4 & \text { if } \mathrm{x} < 2 \\ 3 & \text { if } \mathrm{x} > 2\end{array} \quad\right.$ as is shown in Fig. 20.

Show $\lim\limits_{\mathrm{x} \rightarrow 2} \mathrm{f}(\mathrm{x})$ does not exist.

21. $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{x} & \text { if } \mathrm{x} < 2 \\ 6-\mathrm{x} & \text { if } \mathrm{x} > 2\end{array}\right.$. Show $\lim\limits_{\mathrm{x} \rightarrow 2} \mathrm{f}(\mathrm{x})$ does not exist.

23. Prove: If $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{L}$ and $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{g}(\mathrm{x})=\mathrm{M}$, then $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})=\mathrm{L}-\mathrm{M}$.

1. (a) If $x$ is within $\mathbf{0. 5}$ unit of 3.

(b) If $x$ is within $\mathbf{0. 3}$ unit of 3.

(c) If $\mathrm{x}$ is within $\mathbf{0. 0 2}$ unit of 3.

(d) If $x$ is within $\varepsilon / 2$ unit of 3.

3. (a) If $x$ is within $0.25$ unit of 2.

(b) If $x$ is within $0.1$ unit of 2.

(c) If $\mathrm{x}$ is within $\mathbf{0. 0 2}$ unit of 2.

(d) If $\mathrm{x}$ is within $\varepsilon / 4$ unit of 2.

5. Problem 1: slope $=2, \delta=\varepsilon / 2.$ Problem 3: slope $=4, \delta=\varepsilon / 4$.

General pattern: $\delta=\varepsilon /$ slope for linear functions

7. Each board must be within $0.06 / 3=0.02$ inches of 10 inches in length.

(b) $1.995824623 < x < 2.004158016$

9. (a) $1.957433821 < x < 2.040827551$ (b) $1.995824623 < \mathrm{x} < 2.004158016$

11. (a) $0 < x < 8$ (b) $2.99920004 < x < 3.00080004$

13. Each piece of wire must be within $0.005996404$ inches of 5 inches.

15. & 17. See Figures

19. Take $\varepsilon=1 / 2$ (or smaller).

If $x>2$ and If $(x)-L \mid < \varepsilon=1 / 2$ then $|2-L| < 1 / 2$ so $3 / 2 < L < 5 / 2$.

If $\mathrm{x} < 2$ and $\mid \mathrm{f}(\mathrm{x})-L \mathrm{k} < \varepsilon=1 / 2$ then $\mid 3-L \mathrm{~K} < 1 / 2$ so $5 / 2 < L < 7 / 2$.

There is no value of $\mathrm{L}$ that is both larger than $5 / 2$ and smaller than $5 / 2$ so the limit does not exist.

21. Take $\varepsilon=1 / 2$ (or smaller) and suppose $\mathrm{x}$ is within 1 of $2(1 < \mathrm{x} < 3)$.

If $1 < \mathrm{x} < 2$ and $|\mathrm{f}(\mathrm{x})-L|=|\mathrm{x}-L|=|L-x| < \varepsilon=1 / 2$ then $-1 / 2 < L-x < 1 / 2$

so $x-1 / 2 < L < x+1 / 2$ and $L < 2.5$.

If $2 < x < 3$ and $|f(x)-L|=|f(x)-L|=|L-6+x| < \varepsilon=1 / 2$ then $-1 / 2 < L-6+x < 1 / 2$

so $5.5 < L+x < 7.5$ and $2.5 < L$.

There is no value of $\mathrm{L}$ that is both larger than $2.5$ and smaller than $2.5$ so the limit does not exist.

23. This proof is very similar to the proof of the second theorem on page 9.

Assume that $\lim _{x \rightarrow a} f(x)=L$ and $\lim _{x \rightarrow a} g(x)=M$. Then, given any $\varepsilon>0$, we know $\varepsilon / 2>0$ and that there are deltas for $\mathrm{f}$ and $\mathrm{g}, \delta_{\mathrm{f}}$ and $\delta_{\mathrm{g}}$, so that

if $|x-a| < \delta_{f}$, then $|f(x)-L| < \varepsilon / 2$ ("if $x$ is within $\delta_{f}$ of a, then $f(x)$ is within $\varepsilon / 2$ of $L$", and

if $|x-a| < \delta_{g}$, then $|g(x)-M| < \varepsilon / 2$ ("if $x$ is within $\delta_{g}$ of $a$, then $g(x)$ is within $\varepsilon / 2$ of M").

Let $\delta$ be the smaller of $\delta_{f}$ and $\delta_{g}$. If $|x-a| < \delta$ then $|f(x)-L| < \varepsilon / 2$ and $|g(x)-M| < \varepsilon / 2$

so $\mid(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x}))-(\mathrm{L}-\mathrm{M}))|=|(\mathrm{f}(\mathrm{x})-\mathrm{L})+(\mathrm{M}-\mathrm{g}(\mathrm{x})) \mid$ (rearranging the terms)

$\leq|f(x)-L|+|M-g(x)| \quad$ (by the Triangle Inequality for absolute values)

$< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \quad$ (by the definition of the limits for $\mathrm{f}$ and $\mathrm{g}$).