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
