What is the limits?

$\lim \left(x \to 2\right)$ $\left(\sqrt{x + 2} + {x}^{2} - 6\right)$/$\left(x - 2\right)$

Mar 27, 2018

The limit equals $\frac{17}{4}$

Explanation:

Use l'Hospitals rule, because we're in the $\frac{0}{0}$ form.

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

$L = \frac{1}{2 \sqrt{4}} + 4$

$L = \frac{1}{4} + 4 = \frac{17}{4}$

The graph of the function confirms. Hopefully this helps!

Mar 27, 2018

$\frac{17}{4}$

Explanation:

Here,
$L = {\lim}_{x \to 2} \frac{\sqrt{x + 2} + {x}^{2} - 6}{x - 2}$

L=lim_(x to2)((x^2-6)+sqrt(x+2))/(x-2)xx((x^2-6)- sqrt(x+2))/((x^2-6)-sqrt(x+2))

$= {\lim}_{x \to 2} \frac{{\left({x}^{2} - 6\right)}^{2} - {\left(\sqrt{x + 2}\right)}^{2}}{\left(x - 2\right) \left(\left({x}^{2} - 6\right) - \sqrt{x + 2}\right)}$

=lim_(x to2)1/((x^2-6)-sqrt(x+2))xxlim_(x to2)(x^4-12x^2+36-x- 2)/(x-2)

$= \frac{1}{4 - 6 - 2} \times {\lim}_{x \to 2} \frac{{x}^{4} - 4 {x}^{2} - 8 {x}^{2} + 16 x - 17 x + 34}{x - 2}$

$= - \frac{1}{4} {\lim}_{x \to 2} \frac{{x}^{2} \left({x}^{2} - 4\right) - 8 x \left(x - 2\right) - 17 \left(x - 2\right)}{x - 2}$

$= - \frac{1}{4} {\lim}_{x \to 2} \left[{x}^{2} \left(x + 2\right) - 8 x - 17\right]$

$= - \frac{1}{4} \left(4 \left(4\right) - 16 - 17\right)$

$= \frac{17}{4}$