How to Find the Limit of (tanx)^cosx as x Approaches Infinity?

fk378
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Homework Statement


Find the limit of (tanx)^cosx as x-->infinity
Rearrange the equation so that you can use L'Hopital's rule for the form of (infinity/infinity)

The Attempt at a Solution


I did ln(tanx)^cosx = cosxlntanx
I know the limit of tanx as x-->infinity is pi/2
the limit of cosx as x-->infinity is infinity

Now, I don't know where to go from here
 
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"the limit of cosx as x-->infinity is 1 (or is it infinity?)"
That limit doesn't exist.
 
Oh okay. So now I have infinity x infinity. I can use L'Hopital's rule but I don't know how to set up the function.
 
No, you cannot use L'Hopital's rule for that. "Infinity* infinity" is not one of cases for which you can use L'Hopital's rule- nor do you need to. You have already been told the answer.
 
I just noticed in the directions it says to rearrange the problem so that you can use L'Hopital's rule.
 
fk378 said:
I just noticed in the directions it says to rearrange the problem so that you can use L'Hopital's rule.

Well then, what are the special cases in which L'Hopital's Rule can be applied? That is what are indeterminate forms?

Casey
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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