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

Question

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

Elijah Abram · Accepted Answer

The limit has value #2#.

L'Hospital's Rule states for #lim_(x->a) (f(x))/(g(x)) = lim_(x-> a) (f'(x))/(g'(x))# if and only if #f(a) = g(a) =0# Checking, the limit is currently of the form #0/0#, so we can use l'Hospital's Rule. The derivative of #t^2# is #2t#. The derivative of #e^t - t - 1# is #e^t - 1#. The limit becomes #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 #2t# is #2# and the derivative of #e^t - 1# is #e^t#. The limit becomes #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!

Adam Adkins · Answer

undefined To find the limit of $ \frac{t^2}{e^t - t - 1} $ as $ t $ approaches $ 0 $ using L'Hôpital's Rule, we first check if the limit is of the form $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $. In this case, as $ t $ approaches $ 0 $, both the numerator and the denominator approach $ 0 $, so it is of the form $ \frac{0}{0} $.

Now, we differentiate both the numerator and the denominator with respect to $ t $ separately.

Derivative of numerator:
$ \frac{d}{dt}(t^2) = 2t $

Derivative of denominator:
$ \frac{d}{dt}(e^t - t - 1) = e^t - 1 $

Now, we re-evaluate the limit using the derivatives:

$ \lim_{t \to 0} \frac{2t}{e^t - 1} $

Again, as $ t $ approaches $ 0 $, both the numerator and the denominator approach $ 0 $, so we can apply L'Hôpital's Rule again.

Differentiating both the numerator and the denominator:

Derivative of numerator:
$ \frac{d}{dt}(2t) = 2 $

Derivative of denominator:
$ \frac{d}{dt}(e^t - 1) = e^t $

Now, we re-evaluate the limit using the derivatives:

$ \lim_{t \to 0} \frac{2}{e^t} $

As $ t $ approaches $ 0 $, $ e^t $ approaches $ e^0 = 1 $, so the limit is $ \frac{2}{1} = 2 $. Therefore, the limit of $ \frac{t^2}{e^t - t - 1} $ as $ t $ approaches $ 0 $ is $ 2 $.

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

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