# lim_(xrarr-3)(x+3)/(4-sqrt(2x+22)) by using L'Hôpital's Rule?

Aug 26, 2017

The limit equals $- 4$.

#### Explanation:

If we try evaluating into the initial expression, we get

$L = \frac{0}{0} = \infty$

So we can use l'Hopitals Rule.

$L = {\lim}_{x \to - 3} \frac{1}{- \frac{2}{2 \sqrt{2 x + 22}}}$

L = lim_(x->-3) 1/(- 1/sqrt((2x + 22))

$L = {\lim}_{x \to - 3} - \sqrt{2 x + 22}$

Now we can evaluate directly:

$L = - \sqrt{2 \left(- 3\right) + 22}$

$L = - \sqrt{16}$

$L = - 4$

If we do a graphical check, we obtain the same result.

graph{(x + 3)/(4 - sqrt(2x + 22)) [-10, 10, -5, 5]}

Hopefully this helps!