# How do you find lim (x^3+4x+8)/(2x^3-2) as x->1^+ using l'Hospital's Rule or otherwise?

Feb 18, 2017

The limit does not exist (it is infinite).

#### Explanation:

L'Hôpital's rule cannot be used as the limit is not of an indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$

The limit does not exist (it is infinite).

This is the graph of the function $\frac{{x}^{3} + 4 x + 4}{2 {x}^{3} - 2}$

graph{(x^3+4x+4)/(2x^3-2) [-10, 10, -25, 25]}

As $x \rightarrow {1}^{+} \implies \frac{{x}^{3} + 4 x + 4}{2 {x}^{3} - 2} \rightarrow \infty$