next up previous
Next: About this document ...

Math 112 Solutions For Section 2.3, p. 132

1.
(b) 1. (d) DNE. (f) $ \frac{3}{4}$. (h) DNE. (j) $ -4$. (k) $ \frac{3}{2}$.
2.
(b) $ -\frac{1}{2}$. (d) DNE. (e) 0. (f) $ \frac{1}{4\sqrt 2}$. (g) $ \frac{1}{2\sqrt 2}$. (h) $ \frac{1}{6}$. (k) 1. (l) $ \frac{2}{3}$.

E.
If we substitute in $ x=1$ in the top and the bottom, we get the $ 0/0$ case. So we look for a factor of $ x-1$ in the top and in the bottom. Since $ x^2-1=(x+1)(x-1)$ and $ x^3-1=(x-1)(x^2+x+1)$ (use polynomial or synthetic division to divide $ x^3-1$ by $ x-1$), we can cancel the $ x-1$ from the top and from the bottom. Thus, $ \lim_{x\to1}\frac{x^2-1}{x^3-1}=\lim_{x\to1}\frac{(x+1)(x-1)}{(x-1)(x^2+x+1)}=
\lim_{x\to1}\frac{x+1}{x^2+x+1}=\frac{2}{3}$.

3.
(a) $ \sqrt 2$. (d) $ \frac{2}{\sqrt 3}$. (e) $ \frac{2}{3}$. (f) 2. (g) $ \frac{2}{3}$. (h) (l) DNE. $ \frac{2}{3}$. (k) 0.
4.
(a) 1. (b) 0. (d) 0. (e) 0. (h) DNE. (i) 0. (j) DNE (undefined for $ x>\pi$).
5.
Theorem. $ \displaystyle{\lim_{x\to c}f(x)=L \Longleftrightarrow
\lim_{x\to c}\vert f(x)-L\vert\vert=0}$.

Proof ( $ \Rightarrow$) If $ f(x)\to L$ as $ x\to c$, then $ 0<\vert x-c\vert<\delta
\Rightarrow \vert f(x)-L\vert<\epsilon \Rightarrow \vert\vert f(x)-L\vert-0\vert<\epsilon$. Hence $ \vert f(x)-L\vert\to 0$ as $ x\to c$.

( $ \Leftarrow$) If $ \vert f(x)-L\vert\to 0$ as $ x\to c$, then $ 0<\vert x-c\vert<\delta
\Rightarrow \vert\vert f(x)-L\vert-0\vert<\epsilon \Rightarrow \vert f(x)-L\vert<\epsilon$. Therefore $ f(x)\to L$ as $ x\to c$.

6.
For example, $ f(x)=\dfrac{\vert x\vert}{x}$ has no limit as $ x\to 0$, but $ \lim_{x\to 0} \left\vert\dfrac{\vert x\vert}{x}\right\vert=1$.

8.
Theorem. If $ \displaystyle{\lim_{x\to c}f(x)=L\neq0\text{ and
}\lim_{x\to c}g(x)=0,\text{ then }\lim_{x\to c}f(x)g(x)=0}$.

Proof. Since $ f(x)\to L$ as $ x\to c$, we can choose $ x$ close enough to $ c$ that $ \vert f(x)\vert<L+1$. Then, given any $ \epsilon>0$, we can choose $ x$ close enought to $ c$ that $ \vert g(x)\vert<\frac{\epsilon}{L+1}$. Then, with $ x$ close enough to $ c$, we have $ \vert f(x)g(x)-0\vert=\vert f(x)\vert\vert g(x)\vert<(L+1)\cdot\frac{\epsilon}{L+1}=\epsilon$.

14.
(a) 0 (b) 0 (c) 0

(d) Undetermined. For example, if $ f(x)=\sin x$ and $ g(x)=\sin^{2}ax$, then $ \lim_{x\to 0}\dfrac{g(x)}{xf(x)}=a^{2}$, which depends on $ a$.

(e) Undetermined. For example, if $ g(x)=\sin^{2}ax$, then $ \lim_{x\to
0}\dfrac{g(x)}{x^{2}}=a^{2}$, which depends on $ a$.

16.
$ \sin x=\sin(\frac{x-c}{2}+\frac{x+c}{2})=\sin\frac{x-c}{2}
\cos\frac{x+c}{2}+\cos\frac{x-c}{2}\sin\frac{x+c}{2}$, and

$ \sin c=\sin(\frac{c-x}{2}+\frac{c+x}{2})=-\sin\frac{x-c}{2}\cos\frac{x+c}{2}
+\cos\frac{x-c}{2}\sin\frac{x+c}{2}$. Hence

$ \sin x-\sin c=2\sin\frac{x-c}{2}
\cos\frac{x+c}{2}$.

18.
(a) bounded, $ B=1$. (b) not bounded. (c) bounded, $ B=1.56$. (d) not bounded. (e) bounded, $ B=1$. (f) bounded, $ B=1$.

19.
$ f(x)=x$ is not bounded, but $ \frac{f(x)}{x}=1$ is bounded for $ x$ not 0. If $ \lim_{x\to 0}f(x)$ exists and is some nonzero number, then $ \frac{f(x)}{x}$ is not bounded near 0. Hence if $ \frac{f(x)}{x}$ is bounded for all $ x$, then either $ f(x)\to 0$ as $ x\to 0$ or $ \lim_{x\to 0}f(x)$ does not exist.

20.
(a) $ s=2r\sin\frac{\pi}{n}$. (b) $ P=2rn\sin\frac{\pi}{n}$. (c) $ C=\lim_{n\to\infty}2rn\sin\frac{\pi}{n}=2\pi r\lim_{n\to\infty}\dfrac{\sin
\pi/n}{\pi/n}=2\pi r\cdot 1=2\pi r$. (d) The area of an inscribed polygon with n sides is $ A_{n}=nr^{2}\sin\frac{\pi}{n}\cos\frac{\pi}{n}$ and the area of the circle is $ A=\lim_{n\to\infty}A_{n}=\pi
r^{2}\lim_{n\to\infty}\frac{\sin \pi/n}{\pi/n}\cos\frac{\pi}{n}=\pi r^{2}$.



next up previous
Next: About this document ...
Jason Grout 2003-01-29