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2.3

Suppose $c \in {\cal R}$ and suppose \hfil\break
$lim_{x \rightarrow a} f(x)$ and $lim_{x \rightarrow a} g(x)$ exist.
Then

$lim_{x \rightarrow a} [f(x) + g(x)] $ = $lim_{x \rightarrow a} f(x)$ + $lim_{x \rightarrow a} g(x)$ 

$lim_{x \rightarrow a} [cf(x)] = $ $c ~lim_{x \rightarrow a} f(x)$ 

$lim_{x \rightarrow a} [f(x)g(x)] = $ $lim_{x \rightarrow a} f(x)$ $lim_{x \rightarrow a} g(x)$ 

$lim_{x \rightarrow a} {f(x) \over g(x)} = $ 
${lim_{x \rightarrow a} f(x) \over lim_{x \rightarrow a} g(x)}$ if 
$lim_{x \rightarrow a} g(x) \not= 0$  

Defn:  $f$ is continuous at $a$ \hfil \break
iff $lim_{x \rightarrow a} f(x) = f(lim_{x \rightarrow a} x) = $ 

If $f$ is continuous then 
 
\centerline{$lim_{x \rightarrow a} f(g(x)) = f(lim_{x \rightarrow a}g(x))$ }
\eject

Theorem:  If $f(x) \leq g(x)$ near $a$ (except possibly at $a$) and if $lim_{x \rightarrow a} f(x)$ and $lim_{x \rightarrow a} g(x)$ exist, then
$$lim_{x \rightarrow a} f(x) \leq lim_{x \rightarrow a} g(x)$$

\vfill

Squeeze theorem:  \hfil \break
If $f(x) \leq g(x) \leq h(x)$ near $a$ (except possibly at $a$) and if $lim_{x \rightarrow a} f(x) = L$ and $lim_{x \rightarrow a} h(x) = L$, then   

\centerline{$lim_{x \rightarrow a} g(x) = L$}

\vfill

Example: ~~ $g(x) = x sin{1 \over x} $

%%$-|x| \leq x sin{1 \over x} \leq |x|$

%%,  $lim_{x \rightarrow 0} -|x| = 0$,  $lim_{x \rightarrow 0} |x| = 0$.  
%%Hence ,  $lim_{x \rightarrow 0} x sin{1 \over x}  = 0$   

%%\vskip 5pt
%%\hrule

%%\vfill
\eject


Defn:  $lim_{x \rightarrow a} f(x) = L$ if

{ $x$ close to  $a$ (except possibly at $a$) \hfil \break implies $f(x)$ is close to $L$. }

\vfill
\eject

Defn:  $lim_{x \rightarrow a} f(x) = L$ if

{ $x$ close to  $a$ (except possibly at $a$) \hfil \break implies $f(x)$ is close to $L$. }

\vfill
\eject

Defn:  $lim_{x \rightarrow a} f(x) = L$ if \hfil \break for all $\epsilon > 0$, there exists a $\delta > 0$ such that
\hfil \break $0 < |x - a| < \delta$ implies $|f(x) - L| < \epsilon$

Show $lim_{x \rightarrow 1} 2 = $ 




\vfill
\eject
Defn:  $lim_{x \rightarrow a} f(x) = L$ if \hfil \break for all $\epsilon > 0$, there exists a $\delta > 0$ such that
\hfil \break $0 < |x - a| < \delta$ implies $|f(x) - L| < \epsilon$



Show $lim_{x \rightarrow 4} 2x + 3 = $ 

\vfill
\eject


Defn:  $lim_{x \rightarrow a^-} f(x) = L$ if

{ $x$ close to  $a$ and $x < a$ \hfil \break implies $f(x)$ is close to $L$. }

\vfill 
Defn:  $lim_{x \rightarrow a^+} f(x) = L$ if

{ $x$ close to  $a$ and $x > a$ \hfil \break implies $f(x)$ is close to $L$. }
\vfill
\eject

Defn:  $lim_{x \rightarrow a} f(x) = \infty$ if

{ $x$ close to  $a$ (except possibly at $a$) \hfil \break implies $f(x)$ is large. }
\vfill


Defn:  $lim_{x \rightarrow a} f(x) = -\infty$ if

{ $x$ close to  $a$ (except possibly at $a$) \hfil \break implies $f(x)$ is negative and $|f(x)|$ is large. }




\end