Practice Problems

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

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 $\mathrm{g}$).