1. ## 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?

2. 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

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

This limit is not of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ 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

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

3. wait I thought dividing by zero doesn't exist. you're saying 1/0 is infinity?

4. 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.

5. okay I graphed it on my calculator so now it makes sense. is it always a good idea to check like that?

6. 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.

7. thank you both for your help