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

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


Ex:  $lim_{x \rightarrow 9} e^{\sqrt{x} } - 2\sqrt{x} + 4$ 

\eject



$lim_{x \rightarrow 3} {x^2 - 1 \over x + 3}$ 



\vfill



$lim_{x \rightarrow 3} {x^2 - 1 \over x - 3}$ 



\vfill



$lim_{x \rightarrow 3} {(x^2 - 1)(x - 3)\over x - 3}$ 



\vfill





$lim_{x \rightarrow 3} { x - 3 \over x^2 - 1 }$ 



\vfill





$lim_{x \rightarrow 3} {(x-4)^2 \over x^5(x-8)^9(x - 3)^3}$ 



\vfill





$lim_{x \rightarrow 3} {(x-4)^2(x - 3) \over x^5(x-8)^9(x - 3)^3}$ 







\eject



Suppose $f(x) = \sqrt{x}$.  Find $lim_{h \rightarrow 0} {f(x + h) - f(x) \over h}$ where $x > 0$









\end