# L'Hopitals Rule

• Sep 27th 2009, 08:16 AM
mech.engineer.major
L'Hopitals Rule
here's the problem:

Find the limit of lim (e^t)-1 / t^3
t-->0

I thought to use L'Hopital's Rule 3 times to get lim e^t / 6
t-->0

Then it would no longer be indeterminate (0/0) and I would have 1/6 as my answer.

But my book says the answer is infinity.

What did I do wrong?
• Sep 27th 2009, 08:22 AM
TheEmptySet
Quote:

Originally Posted by mech.engineer.major
here's the problem:

Find the limit of lim (e^t)-1 / t^3
t-->0

I thought to use L'Hopital's Rule 3 times to get lim e^t / 6
t-->0

Then it would no longer be indeterminate (0/0) and I would have 1/6 as my answer.

But my book says the answer is infinity.

What did I do wrong?

You need to check the hypothesis each time you use L.H's rule.

After one application you get

$\displaystyle \lim_{t \to 0}\frac{e^{t}-1}{t^3}=\lim_{t \to 0}\frac{e^{t}}{3t^2}$

This limit is not of the form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ so we CANNOT use L.H's rule.

Now if we take the limit the numerator goes to 1 and the denominator
goes to zero from both sides so

$\displaystyle \lim_{t \to 0}\frac{e^{t}-1}{t^3}=\lim_{t \to 0}\frac{e^{t}}{3t^2}=\infty$
• Sep 27th 2009, 08:30 AM
mech.engineer.major
wait I thought dividing by zero doesn't exist. you're saying 1/0 is infinity?
• Sep 27th 2009, 08:31 AM
Prove It
Quote:

Originally Posted by mech.engineer.major
wait I thought dividing by zero doesn't exist. you're saying 1/0 is infinity?

You can't divide by 0, but you can see what happens when a function has a denominator that TENDS to 0.
• Sep 27th 2009, 08:35 AM
mech.engineer.major
okay I graphed it on my calculator so now it makes sense. is it always a good idea to check like that?
• Sep 27th 2009, 08:37 AM
Prove It
Quote:

Originally Posted by mech.engineer.major
okay I graphed it on my calculator so now it makes sense. is it always a good idea to check like that?

Absolutely.
• Sep 27th 2009, 08:40 AM
mech.engineer.major
thank you both for your help