# How do you find the limit of  ( sqrt(x^2+11) - 6 ) / (x^2 - 25) as x approaches 5?

Jan 25, 2016

$\frac{1}{12}$

#### Explanation:

Notice how evaluating the limit presently would be impossible, since the denominator of the fraction would be $0$.

A good strategy when trying to deal with expression with radicals and other general nastiness is to multiply by a conjugate. We can try to multiply this by the conjugate of the numerator.

$= {\lim}_{x \rightarrow 5} \frac{\sqrt{{x}^{2} + 11} - 6}{{x}^{2} - 25} \left(\frac{\sqrt{{x}^{2} + 11} + 6}{\sqrt{{x}^{2} + 11} + 6}\right)$

In the numerator, we can recognize that this is really the difference of squares pattern in reverse, where $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$.

$= {\lim}_{x \rightarrow 5} \frac{{\left(\sqrt{{x}^{2} + 11}\right)}^{2} - {6}^{2}}{\left({x}^{2} - 25\right) \left(\sqrt{{x}^{2} + 11} + 6\right)}$

$= {\lim}_{x \rightarrow 5} \frac{{x}^{2} + 11 - 36}{\left({x}^{2} - 25\right) \left(\sqrt{{x}^{2} + 11} + 6\right)}$

$= {\lim}_{x \rightarrow 5} \frac{{x}^{2} - 25}{\left({x}^{2} - 25\right) \left(\sqrt{{x}^{2} + 11} + 6\right)}$

Notice the ${x}^{2} - 25$ terms in the numerator and denominator will cancel.

$= {\lim}_{x \rightarrow 5} \frac{1}{\sqrt{{x}^{2} + 11} + 6}$

We can now evaluate the limit by plugging in $5$ for $x$.

$= \frac{1}{\sqrt{25 + 11} + 6} = \frac{1}{\sqrt{36} + 6} = \frac{1}{12}$

We can check a graph of the original function:

graph{( sqrt(x^2+11) - 6 ) / (x^2 - 25) [-1.53, 12.52, -0.05, .15]}

Even though the point at $x = 5$ is undefined, it is very close to $\frac{1}{12} = 0.08 \overline{3}$.