Question

Find the following limit.
$\underset{x\to -1}{lim}\frac{x^3-2x-1}{x^5-2x-1}.$

Answer 1

$\underset{x\to -1}{lim}\frac{x^3-2x-1}{x^5-2x-1}$

here form of $\frac{0}{0}$ So applying L Hospital Rule:

$\underset{x\to -1}{lim}\frac{\frac{d}{dx}(x^3-2x-1)}{\frac{4}{dx}(x^4-2x-1)}$

$\underset{x\to -1}{lim}\frac{3x^2-2}{5x^4-2}$

$=\frac{3(-1{)}^2-2}{5(-1{)}^{(}4)-2}=\frac{3\left(1\right)-2}{5\left(1\right)-2}$

$=\frac{1}{3}$.