# What is #lim_(x->oo) e^x/(x-1)# ?

##### 2 Answers

#### Explanation:

If we evaluate this by direct substitution we get

This answer is not helpful. Since

L'Hospital's rule provides a solution for us. L'Hospital's rule states that:

In other words, **if and only if** a limit ends up as **not the quotient rule**. You are to take the derivative of the top and the bottom separately.

Applying the rule,

#### Explanation:

The simple answer is that

Hence:

#lim_(x->oo) e^x/(x-1) = oo#

A little more formally:

#e^x = sum_(k=0)^oo x^k/(k!) > x^2/2#

Hence:

#lim_(x->oo) e^x/(x-1) >= lim_(x->oo) (x^2/2)/(x-1) >= lim_(x->oo) 1/2(x^2/x) = lim_(x->oo) x/2 = oo#