# Question of the existence of a limit using l'Hopital's Rule

• Dec 18th 2009, 03:22 PM
Bryn
Question of the existence of a limit using l'Hopital's Rule
Hi,

The problem in question is as follows:

Limit as u tends to 0 of:

[(u+tan2u)/(u-tanu)]

When differentiating the denominator it becomes 1-sec^2 u which makes the denominator 0. I was under the impression that l'hopitals rule doesn't work under this condition of g1(x) being equal to 0.

That being said, the answer sheet I have continues to the conclusion that it tends to infinity disregarding what I have seen. Am I mistaken

Many thanks
• Dec 18th 2009, 03:30 PM
mr fantastic
Quote:

Originally Posted by Bryn
Hi,

The problem in question is as follows:

Limit as u tends to 0 of:

[(u+tan2u)/(u-tanu)]

When differentiating the denominator it becomes 1-sec^2 u which makes the denominator 0. I was under the impression that l'hopitals rule doesn't work under this condition of g1(x) being equal to 0.

That being said, the answer sheet I have continues to the conclusion that it tends to infinity disregarding what I have seen. Am I mistaken

Many thanks

Applying l'Hopital's Rule in the usual way gives $\lim_{u \to 0} \frac{1 + 2 \sec^2 (2u)}{1 - \sec^2 (u)}$ and the answer given by the book should now be obvious.
• Dec 18th 2009, 03:32 PM
11rdc11
Quote:

Originally Posted by Bryn
Hi,

The problem in question is as follows:

Limit as u tends to 0 of:

[(u+tan2u)/(u-tanu)]

When differentiating the denominator it becomes 1-sec^2 u which makes the denominator 0. I was under the impression that l'hopitals rule doesn't work under this condition of g1(x) being equal to 0.

That being said, the answer sheet I have continues to the conclusion that it tends to infinity disregarding what I have seen. Am I mistaken

Many thanks

L'hopital rule does work under this situation.

$\frac{3}{0} = \infty$
• Dec 18th 2009, 04:16 PM
Bryn
I understand the answer and the process, my question is about the restriction of when you can use L'hopital's rule. I believe you that it works here but for the reasons I said, I don't understand why. Th first order derivative function of the denominator =0 and I thought this was a condition for not using this rule. Am I mistaken?
• Dec 18th 2009, 04:32 PM
mr fantastic
Quote:

Originally Posted by Bryn
I understand the answer and the process, my question is about the restriction of when you can use L'hopital's rule. I believe you that it works here but for the reasons I said, I don't understand why. Th first order derivative function of the denominator =0 and I thought this was a condition for not using this rule. Am I mistaken?

L'Hopital's Rule is used on indeterminant forms like 0/0, which you clearly have here. I don't see what the trouble is.
• Dec 19th 2009, 09:14 PM
oblixps
the only time L'Hospital's Rule fails is when your limit is not an indeterminate form or when you take the limit of the quotient of the derivatives, the limit does not exist. if the limit exists, or if the limit goes to positive of negative infinite, L'Hospital's Rule is valid. so if you take $\lim_{u \to 0} \frac{1 + 2 \sec^2 (2u)}{1 - \sec^2 (u)}$ you should get - $\infty$