Question

Examine the following function for continuity.

$f\left(x\right)=$ $\frac{x^2-25}{x+5}$, $x$ $\ne$ $5$

Answer 1

The given function is $f\left(x\right)=\frac{x^2-25}{x+5},x\ne 5$

For any real number $c\ne -5,$ we obtain

$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}\frac{(x+5)(x-5)}{x+5}=\underset{x\to c}{lim}(x-5)=(c-5)$

Also, $f\left(c\right)=\frac{(c+5)(c-5)}{(c+5)}=(c-5)$(as c $\ne$ -5)

$\therefore$ $\underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$

Hence, $f$ is continuous at every point in the domain of $f$ and therefore, it is a continuous function.