# How do you find lim_(x to 2) (x^3-6x-2)/(x^3+4), using l'Hospital's Rule or otherwise?

Nov 17, 2017

Because the expression evaluated at $x = 2$ does not yield an indeterminate form, (e.g. $\frac{0}{0}$ or $\frac{\infty}{\infty}$), one cannot use L'Hospital's rule. The limit is merely the expression evaluated at $x = 2$.

#### Explanation:

Evaluate the expression at $x = 2$:

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

The limit is:

${\lim}_{x \to 2} \frac{{x}^{3} - 6 x - 2}{{x}^{3} + 4} = - \frac{1}{2}$