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!