My teacher gave me this problem as an honors problem. She said that at one point I will have to use L'Hopitals rule to solve, but I have no idea where to start.

My teacher gave me this problem as an honors problem. She said that at one point I will have to use L'Hopitals rule to solve, but I have no idea where to start.

Let \(\displaystyle f=x+1 \) and \(\displaystyle g=\cot x \).

We end up with a limit of the form \(\displaystyle 1^\infty \). To fix this notice \(\displaystyle \displaystyle f^g = \exp\left(\frac{\ln f}{1/g}\right) \).

So \(\displaystyle \displaystyle \lim_{x\to0}f^g = \exp\left(\lim_{x\to0}\frac{\ln f}{1/g}\right) \) and the limit is of the form \(\displaystyle \displaystyle \frac00 \) (can you see why?).

I cannot thank you enough! I did not see that it was f raised to the g. My only other question is on the second line you have f^g= exp(ln f/(1/g)). What is the exp part?