# How do you find #lim t^2/(e^t-t-1)# as #t->0# using l'Hospital's Rule?

##### 1 Answer

Feb 27, 2017

The limit has value

#### Explanation:

L'Hospital's Rule states for

Checking, the limit is currently of the form

The derivative of

#lim_(t->0) (2t)/(e^t - 1)#

#(2(0))/(e^0 - 1)#

#0/0#

We are in the same position as we were originally. We will once again apply l'Hospitals. The derivative of

#lim_(t->0) 2/e^t#

#2/e^0#

#2/1#

#2#

A graphical verification yields the same result.

graph{x^2/(e^x - x - 1) [-10, 10, -5, 5]}

Hopefully this helps!