# 5 c) Examine the following functions for continuity.$f (x) = \frac{x ^2-25}{x+5}, x \neq -5$

$f(x ) = \frac{x^2-25}{x+5}$
For every real number  k , $k \neq -5$
$f(k) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = \frac{k^2-5^2}{k+5}= \frac{(k +5)(k-5)}{k+5} = k-5\\ \lim_{x\rightarrow k}f(x ) = f(k)$
Hence, function  $f(x ) = \frac{x^2-25}{x+5}$ continuous for every real value of x , $x \neq -5$