Calculus 2 limits as x-->0+

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August 3rd 2010, 05:22 PM
JubJub1991

Calculus 2 limits as x-->0+

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.

1. Evaluate the limit, if it exists.

l i m (x+1)^cotx
n-->0+

Any help would be greatly appreciated
August 3rd 2010, 05:30 PM
chiph588@

Quote:

Originally Posted by JubJub1991

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.

1. Evaluate the limit, if it exists.

l i m (x+1)^cotx
n-->0+

Any help would be greatly appreciated

Let $f=x+1$ and $g=\cot x$ .

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

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

Now apply L'Hopital's rule.
August 3rd 2010, 05:55 PM
JubJub1991

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?
August 3rd 2010, 06:28 PM
chiph588@

$\exp(x) = e^x$