# How do you evaluate the limit (x^2+2x-8)/(sqrt(x^2+5)-(x+1)) as x approaches 2?

Aug 5, 2016

-18

#### Explanation:

${\lim}_{x \to 2} \frac{{x}^{2} + 2 x - 8}{\sqrt{{x}^{2} + 5} - \left(x + 1\right)}$

$= {\lim}_{x \to 2} \frac{\left(x + 4\right) \left(x - 2\right)}{\sqrt{{x}^{2} + 5} - \left(x + 1\right)}$

let $x = 2 + h , 0 < \left\mid h \right\mid \text{<<} 0$

so the limit becomes

$= {\lim}_{h \to 0} \frac{\left(6 + h\right) \left(h\right)}{\sqrt{9 + 4 h + {h}^{2}} - \left(3 + h\right)}$

by Binomial expansion

$\sqrt{9 + 4 h + {h}^{2}}$

$= 3 {\left(1 + \frac{1}{9} \left(4 h + {h}^{2}\right)\right)}^{\frac{1}{2}}$

$= 3 \left(1 + \frac{1}{2} \cdot \frac{1}{9} \left(4 h + {h}^{2}\right) + O \left({h}^{2}\right)\right)$, remembering that $0 < \left\mid h \right\mid \text{<<} 0$

$= 3 + \frac{2}{3} h + O \left({h}^{2}\right)$

$\implies {\lim}_{h \to 0} \frac{\left(6 + h\right) \left(h\right)}{3 + \frac{2}{3} h + O \left({h}^{2}\right) - \left(3 + h\right)}$

$= {\lim}_{h \to 0} \frac{\left(6 + h\right) \left(h\right)}{- \frac{1}{3} h + O \left({h}^{2}\right)}$

$= {\lim}_{h \to 0} - \frac{3 \left(6 + h\right)}{1 + O \left(h\right)}$

$= - 18$