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

##### 1 Answer

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

The limit does not exist (it is infinite).

This is the graph of the function

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

As