
\documentclass[12pt]{amsart}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage[mathscr]{eucal}
\usepackage[all]{xy}
\usepackage{color}
\def\dsp{\displaystyle}


\title{Limit and Continuity}

\begin{document}

\maketitle
\begin{enumerate}
\item[Solutions:]

\item [1.]For every $\epsilon > 0 $ , there exists $\delta
> 0$ such that

\

$\mid f(x)-\it l \mid < \epsilon$ for all x with $ 0 < \mid x - a
\mid < \delta. $

\

\item[] (or For every $\epsilon > 0 $ , there exists a
corresponding $\delta > 0$ such that for all x ,$  $ $ 0 < \mid x
- a \mid < \delta \Rightarrow $ $ \mid f(x) - \it l \mid <
\epsilon.$ )


\

\item [2.]Arbitrarily give $\epsilon > 0 $.

\

$\dsp \lim_{x \rightarrow a} f(x)= \it l $

\item [] $\Rightarrow$ there exists $\delta_1  > 0$  such that

($0 < \mid x - a \mid < \delta_1$
 $ \Rightarrow \mid f(x) - \it
l \mid < \frac{\epsilon}{2}$$) \quad \quad \quad \ldots (1)$

\

$\dsp \lim_{x \rightarrow a} g(x)= \it m $

\item [] $\Rightarrow$ there exists $\delta_2  > 0$  such that

($0 < \mid x - a \mid < \delta_2$
 $ \Rightarrow \mid g(x) - \it
m \mid < \frac{\epsilon}{2}$$) \quad \quad \quad \ldots (2)$

\

\item [] Take $\delta = \min \{\delta_1,\delta_2\} $ , then

\

$0 < \mid x - a \mid < \delta $

\

\item [] $ \Rightarrow$
$\left\{ \begin{array}{ll}
           0 < \mid x - a \mid < \delta_1 \\
           0 < \mid x - a \mid < \delta_2
        \end{array} \right.$

\

\

\item [] $ \Rightarrow$
$\left\{ \begin{array}{ll}
           \mid f(x) - \it l $ $\mid < \frac{\epsilon}{2} \quad \quad ($ ${}^\cdot{}_\cdot{}^\cdot (1)$ $) \\
           \mid g(x) - \it m $$\mid < \frac{\epsilon}{2} \quad \quad ($ ${}^\cdot{}_\cdot{}^\cdot (2)$ $)
        \end{array} \right.$

        \

        \

\item [] $\Rightarrow \mid ( f(x) + g(x) - (\it l+m ) \mid $

\

=$\mid ( f(x)- l ) + ( g(x) - m ) \mid$

\

$\leq \mid f(x)- l \mid + \mid g(x) - m \mid$

\

$<\epsilon$

\

\item [] ${}_\cdot{}^\cdot{}_\cdot$ For an arbitrarily given $\epsilon >
0 $ , there exist $\delta > 0 $ such that 
$$ 0 < \mid x - a \mid < \delta_1 \Rightarrow \mid f(x) + g(x) - (\it l + m ) \mid < \epsilon. $$

\

Hence $\dsp \lim_{x\rightarrow a} \it f(x) + g(x) = l + m $.


\

\

\item [3.]Suppose $g(x) \leq f(x) \leq h(x) $ for all x in some
neighborhood of a , and $ x \neq a . $

\item [] If $\dsp \lim_{x\rightarrow a} g(x)= \it l $ and $\dsp
\lim_{x\rightarrow a}h(x)= \it l, $ then $\dsp \lim_{x\rightarrow a}f(x)= \it l $.

\

\item [4.]$\dsp \lim_{x\rightarrow a}f(x)$ exists, and
$\dsp \lim_{x\rightarrow a}f(x)=f(a)$

\

\item [5.](1) 3 \quad (2) $\frac{1}{2}$ \quad (3) does not exist

\

\item [6.](i) (1)1 \quad (2) 2 \quad (3) does not exists
\item [] (ii) g is not continuous at 0.

${}^\cdot{}_\cdot{}^\cdot \dsp \lim_{x\rightarrow 0}g(x)$ does not
exist.

\

\item [7.](i) (1) 2 \quad (2) 2 \quad (3) 2
\item [] (ii) h is continuous at -1.

 ${}^\cdot{}_\cdot{}^\cdot \dsp \lim_{x\rightarrow
-1}h(x)=h(-1)$

\

\item [8.](i) (1) 0 \quad (2) does not exist
\item [] (ii) 0


\item [9.] $\dsp \lim_{x\rightarrow 0}\sin \frac{1}{x}$ does not
exist.

\

\item []${}^\cdot{}_\cdot{}^\cdot $ When x approaches 0 , the
values of $\sin \frac{1}{x}$ change between \item []\quad -1 and
1 ; they do not approach a fixed number.

\

\item [10.]$-\mid x \mid \leq x\sin(\frac{1}{x}) \leq \mid x \mid $
for all x $ \neq 0 $, and

\

$\left\{ \begin{array}{ll}
          \dsp \lim_{x\rightarrow o }\mid x \mid = 0 \\
           \dsp \lim_{x\rightarrow o }-\mid x \mid = 0
        \end{array} \right.$
\

\item [] Hence , by Sandwich Theorem , $\dsp \lim_{x\rightarrow o } x\sin \frac{1}{x}  = 0 $

\

\item [11.] For every M $>$ 0 , there exists $\delta > 0 $ such
that

\

\item [] $ 0 < \mid x - a \mid < \delta \Rightarrow f(x) > M. $

\end{enumerate}

\end{document}
