Problem

Find the limit. Use l'Hospital's Rule where appro…

04:14

Problem 5 Easy Difficulty

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.
$\lim _{t \rightarrow 0} \frac{e^{t}-1}{t^{3}}$

Answer

$\lim _{x \rightarrow 0} \frac{e^{t}-1}{t^{3}}=\lim _{x \rightarrow 0} \frac{d\left(e^{t}-1\right)}{d\left(t^{3}\right)}=\lim _{x \rightarrow 0} \frac{e^{t}}{3 t^{2}}=\frac{e^{0}}{0^{3}}=\frac{1}{0}=\infty$

Related Courses

Calculus 1 / AB

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Chapter 4

Applications of Derivatives

Section 3

L'Hospital's Rule: Comparing Rates of Growth

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Video Transcript

Okay, so here we have the limit as X is a approaching zero of B to the power of T minus one over t cute. And so first, we will simply e evaluate the limit. Course. I already made a small typo that should be 80. Otherwise, this doesn't make any sense. So first will evaluate the limit, will put t equals zero in there and see what we get. She would get eat. The zero minus 1/0 cubed, which would be one minus 1/0 is 0/0. So therefore Lotito's rule applies. We're gonna take the derivative with respect to t of the top over the bottom and take that limit instead. All right, so the limit of U to the T minus one is going to remain just e to the t and the one has been dropped out. And then the derivative of tea to the power of three will be three t squared. So now we'll try again to simply evaluate the limit At T equals zero. And when we do that, we get eat of the power of zero over three times zero squared, which will give us 1/0. No, no, we have not a 0/0. This means that our women is going to be infinity. This one is getting divided by tinier and tinier numbers, which means the resultant numbers are larger and larger and larger. Eso this limit is equal to infinity.

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