How do I find the limit as x approaches 0 of (x^2)/((e^2x) -1-2x) ? I used L'Hopital's Rule, but keep ending up with 0.

Question

How do I find the limit as x approaches 0 of (x^2)/((e^2x)  -1-2x) ? I used L'Hopital's Rule, but keep ending up with 0.

anonymous · Answer

Well, sometimes the limit is 0. You use L'Hopital's rule to get out of the problem of having numerator and denominator both equal to zero. If you apply the rule and your numerator is still zero, but your denominator is no longer zero, then you've determined that the limit is zero.

anonymous · Answer

Exactly, which is why initially that is precisely what I thought, but I checked the answers (it is a question from a past exam) and apparently it is 1/2...

anonymous · Answer

$$\lim_{x \rightarrow 0}\frac{ x^2 }{ e^{2x}-1-2x }\overset{L}=\lim_{x \rightarrow 0}\frac{ 2x }{ 2e^{2x}-2 }\overset{L}=\lim_{x \rightarrow 0}\frac{ 2 }{4 e^{2x} }$$

anonymous · Answer

Aaaaaaaaaa.... okay, how did you get from step 2 to step 3?

anonymous · Answer

take the derivative again... you can use L'Hopital's so long as it's 0/0 or one of the forms..

anonymous · Answer

WHY DID I NEVER REALIZE THIS. Sorry, my prof is...tutorially challenged. Thank you.

anonymous · Answer

you're welcome