# How do you evaluate the limit (sqrt(x^2-5)+2)/(x-3) as x approaches 3?

Aug 14, 2018

Please go through the Discussion in The Explanation.

#### Explanation:

Observe that,

${\lim}_{x \to 3} \left(\sqrt{{x}^{2} - 5} + 2\right) = \sqrt{{3}^{2} - 5} + 2 = 2 + 2 = 4$.

Hence, as long as $\left(x - 3\right)$ remains in the Denominator , the

limit can not exist.

$\text{However, had the Required Limit been } {\lim}_{x \to 3} \frac{\sqrt{{x}^{2} - 5} - 2}{x - 3}$,

$\text{The Limit} = \lim \frac{\sqrt{{x}^{2} - 5} - 2}{x - 3} \times \frac{\sqrt{{x}^{2} - 5} + 2}{\sqrt{{x}^{2} - 5} + 2}$,

$= \lim \frac{{\left(\sqrt{{x}^{2} - 5}\right)}^{2} - {2}^{2}}{\left(x - 3\right) \left\{\sqrt{{x}^{2} - 5} + 2\right\}}$,

$= \lim \frac{{x}^{2} - 5 - 4}{\left(x - 3\right) \left\{\sqrt{{x}^{2} - 5} + 2\right\}}$,

$= {\lim}_{x \to 3} \frac{\cancel{\left(x - 3\right)} \left(x + 3\right)}{\cancel{\left(x - 3\right)} \left\{\sqrt{{x}^{2} - 5} + 2\right\}}$,

$= \frac{3 + 3}{\sqrt{{3}^{2} - 5} + 2}$,

$= \frac{6}{2 + 2}$,

$= \frac{3}{2}$.