Question

Prove that
$f\left(x\right)=\{\begin{array}{c}\frac{x^2-25}{x-5},whenx\ne 5\\ 10,whenx=5\end{array}$ is continuous at $x=5$.

Answer 1

L.H.L$={lim}_{x\to 5^{-}}\frac{x^2-25}{x-5}$

$={lim}_{x\to 5^{-}}\frac{(x-5)(x+5)}{x-5}$

$={lim}_{x\to 5^{-}}(x+5)$

$=5+5=10$

R.H.L$={lim}_{x\to 5^{+}}\frac{x^2-25}{x-5}$

$={lim}_{x\to 5^{+}}\frac{(x-5)(x+5)}{x-5}$

$={lim}_{x\to 5^{+}}(x+5)$

$=5+5=10$

$\therefore$ L.H.L$=$R.H.L

$f\left(5\right)=10$

$\therefore f\left(x\right)$ is continuous at $x=5$