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Math 112 Solutions for 2.2, p.100

1.
(a) about 1.2. (b) about -1. (c) 0 (d) 0 (e) 0 (f) about $ \frac{1}{2}$. (g) 1 (h) DNE

2.
(a) The limit is 5 because the line $ y=2x-3 $ passes through the point (4, 5). (b) 6. (c) 1. (d) -1. (f) -5. (g) 1. (h) DNE. (j) DNE. (l) 0. (n) 0. (p) DNE.

3.
(a) $ \vert(2x-1)-5\vert<.001\Rightarrow 2\vert x-3\vert<.001\Rightarrow
\vert x-3\vert<.0005$.

(b) The graph of $ y=\frac{1}{x}$ is between .499 and .501 for $ 1.998<x<2.002$.

(c) $ \vert(x^{2}+1)-5\vert=\vert x^{2}-4\vert=\vert x+2\vert\vert x-2\vert$. If $ \vert x-2\vert<1$, then $ x<3$ and $ \vert x+2\vert<5$. If $ \vert x-2\vert<\delta$, then $ \vert x+2\vert\vert x-2\vert<5\delta$, so let $ 5\delta <.001$, or $ \delta < .0002$.

(d) $ \vert\sqrt x-4\vert=\dfrac{\vert x-16\vert}{\vert\sqrt x+4\vert}$. If $ \vert x-16\vert<1$, then $ x>15$ and $ \vert\sqrt x+4\vert>4+\sqrt{15}$. If also $ \vert x-16\vert<\delta$, then $ \vert\sqrt x-4\vert=
\dfrac{\vert x-16\vert}{\vert\sqrt x+4\vert}< \dfrac{\delta}{4+\sqrt{15}}$, so let $ \dfrac{\delta}{4+\sqrt{15}}<.001$. Then $ \delta <(4+\sqrt{15})(.001)
\approx .00787$.

4.
(b) Since $ f(\frac{1}{2})\approx .18877$, plot $ f(x)=\dfrac{x}{e^{x}+1}$ in the window $ [.495, .505]\times [.18777, .18977]$. The graph appears between $ x\approx .4962$ and $ x\approx .5039$. Hence we may choose $ \delta<.003$.

5.
(a) Let $ \delta =.0005$. Then $ \vert x-3\vert<\delta \Rightarrow
\vert(2x-1)-5\vert=2\vert x-3\vert<2\delta=.001$.

(c) $ \vert(x^{2}-1)-8\vert=\vert x^{2}-9\vert=\vert x+3\vert\vert x-3\vert$. If $ \vert x-3\vert<1$, then $ \vert x+3\vert<7$. Let $ \delta =\frac{.004}{7} \approx .0005$. Then $ 0<\vert x-3\vert<\delta \Rightarrow
\vert x-3\vert\vert x+3\vert< (.0005)(7)=.0035<.004$.

6.
(a) Given $ \epsilon > 0$, let $ \delta=\frac{\epsilon}{3}$. Then $ 0
< \vert x-2\vert<\delta \Rightarrow \vert(3x+1)-7\vert=3\vert x-2\vert<3\cdot
\frac{\epsilon}{3}=\epsilon$.

(b) Let $ \epsilon > 0$ be given, and let $ \delta < 6\epsilon$. Then $ \vert x-9\vert<\delta \Rightarrow \vert\sqrt{x}-3\vert=\dfrac{\vert x-9\vert}{\vert\sqrt
x+3\vert}<\dfrac{\delta}{3+\sqrt{9-\delta}}<\dfrac{\delta}{6}<\epsilon$.

(c) $ \vert(x^{2}+4)-5\vert=\vert x^{2}-1\vert=\vert x-1\vert\vert x+1\vert$. If $ \vert x-1\vert<1$, then $ \vert x+3\vert<3$. Hence, given $ \epsilon > 0$, let $ \delta=\min\{1,\frac{\epsilon}{3}\}$. Then $ 0<\vert x-1\vert<\delta \Rightarrow \vert(x^{2}+4)-5\vert=\vert x-1\vert\vert x+1\vert<\frac{\epsilon}{3}\cdot
3=\epsilon$.

(d) $ \vert\frac{2}{x}-(-2)\vert=\frac{2\vert x+1\vert}{\vert x\vert}$. Let $ \vert x+1\vert<\frac{1}{2}$. Then $ -\frac{1}{2}<x+1<\frac{1}{2}\Rightarrow -\frac{3}{2}<x<-\frac{1}{2}$, so $ \vert x\vert>\frac{1}{2}$ and $ \frac{1}{\vert x\vert}<2$. Now given $ \epsilon > 0$, let $ \delta=\min\{\frac{1}{2},\frac{\epsilon}{4}\}$. Then $ 0<\vert x-(-1)\vert<\delta\Rightarrow
\vert\frac{2}{x}-(-2)\vert=2\vert x+1\vert\cdot\frac{1}{\vert x\vert}<2\cdot\frac{\epsilon}{4}\cdot
2=\epsilon$.

11.
Suppose both $ f(x)\to L$ and $ f(x)\to M$ as $ x\to c$, with $ L\neq M$. Let $ \epsilon =\frac{1}{4}\vert L-M\vert$. Then there exists $ \delta_{1}$ such that $ \vert x-c\vert<\delta_{1}\Rightarrow \vert f(x)-L\vert<\epsilon$ and there exists $ \delta_{2}$ such that $ \vert x-c\vert<\delta_{2}\Rightarrow \vert f(x)-M\vert<\epsilon$. Hence, if $ \delta=\min\{\delta_{1},\delta_{2}\}$, then $ \vert x-c\vert<\delta\Rightarrow \vert L-M\vert=
\vert[L-f(x)]+[f(x)-M]\vert\leq \vert f(x)-L\vert+\vert f(x)-M\vert<2\epsilon=\frac{1}{2}\vert L-M\vert$, a contradiction.

14.
(a) $ \frac{1}{2}$. (b) $ \frac{1}{2}$. (c) $ \frac{1}{2}$. (d) -1. (e) 1. (f) DNE. (g) DNE. (h) DNE. (i) DNE. (j) 0. (k) 0. (l) 0.

16.
(b), (c), and (d) are true.

17.
For example, if $ x_{1}=1$, then $ x_{2}=1.5,
x_{3}=1.41\overline{6}, x_{4}=1.4142157, x_{5}=1.4142136,
x_{6}=1.4142136,\dots$. $ g(x)=x\Rightarrow x=\pm\sqrt{2}$; $ \sqrt
2\approx1.4142136$.




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Jason Grout 2003-01-27