# Question #a35d7

Jul 1, 2016

${\lim}_{x \to {25}^{+}} \frac{{x}^{2} + 1}{\sqrt{x} - 5} = \infty$

${\lim}_{x \to {25}^{-}} \frac{{x}^{2} + 1}{\sqrt{x} - 5} = - \infty$

#### Explanation:

${\lim}_{x \to 25} \frac{{x}^{2} + 1}{\sqrt{x} - 5}$

$= {\lim}_{x \to 25} \frac{{x}^{2} + 1}{\sqrt{x} - 5} \cdot \frac{\sqrt{x} + 5}{\sqrt{x} + 5}$

$= {\lim}_{x \to 25} \frac{\left({x}^{2} + 1\right) \left(\sqrt{x} + 5\right)}{x - 25}$

if we substitute $x = 25 + \epsilon$ into the limit [ie $\epsilon = x - 25$], with $0 < \left\mid \epsilon \right\mid < < 1$, then it becomes

$= {\lim}_{\epsilon \to 0} \frac{\left({\left(25 + \epsilon\right)}^{2} + 1\right) \left(\sqrt{25 + \epsilon} + 5\right)}{\epsilon}$

For $\epsilon > 0$, this is $+ \infty$

so ${\lim}_{x \to {25}^{+}} \frac{{x}^{2} + 1}{\sqrt{x} - 5} = \infty$

and for $\epsilon < 0$, this is $- \infty$

so ${\lim}_{x \to {25}^{-}} \frac{{x}^{2} + 1}{\sqrt{x} - 5} = - \infty$