# Evaluate the limit of the indeterminate quotient?

## Feb 8, 2017

${\lim}_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = - \frac{1}{\sqrt{7}}$

#### Explanation:

Evaluate the limit:

${\lim}_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x}$

Rationalize the numerator of the function using the identity: $\left(a + b\right) \left(a - b\right) = \left({a}^{2} - {b}^{2}\right)$:

$\frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = \left(\frac{\sqrt{7 - x} - \sqrt{7 + x}}{x}\right) \left(\frac{\sqrt{7 - x} + \sqrt{7 + x}}{\sqrt{7 - x} + \sqrt{7 + x}}\right) = \frac{7 - x - 7 - x}{x \left(\sqrt{7 - x} + \sqrt{7 + x}\right)} = - \frac{2 x}{x \left(\sqrt{7 - x} + \sqrt{7 + x}\right)} = - \frac{2}{\sqrt{7 - x} + \sqrt{7 + x}}$

Now the limit is determinate:

${\lim}_{x \to 0} \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = {\lim}_{x \to 0} - \frac{2}{\sqrt{7 - x} + \sqrt{7 + x}} = - \frac{1}{\sqrt{7}}$

Feb 9, 2017

$- \frac{1}{\sqrt{7}}$

#### Explanation:

Here is another way of solving it. Slightly messy.

L'hopital's Rule:

${\lim}_{x \to 0} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to 0} \frac{{f}^{'} \left(x\right)}{{g}^{'} \left(x\right)}$ only if the expression is indeterminate .

${f}^{'} \left(x\right)$ is simply the derivative of $f \left(x\right)$ with respect to $x$.

Now via L'hopital's Rule (differentiate the numerator and the denominator separately), ${\lim}_{x \to 0} = \frac{\sqrt{7 - x} - \sqrt{7 + x}}{x} = {\lim}_{x \to 0} \frac{- \frac{1}{2} {\left(7 - x\right)}^{- \frac{1}{2}} - \frac{1}{2} {\left(7 + x\right)}^{- \frac{1}{2}}}{1} = - \frac{1}{2} \left(\frac{1}{\sqrt{7 - x}} + \frac{1}{\sqrt{7 + x}}\right)$

Let $x = 0$ and you get $- \frac{1}{\sqrt{7}}$