Michael's question at Yahoo Answers involving L'Hôpital's rule

[MarkFL]

How to Find Lim x -> 0 of Tan(x)^x with the L'Hopital Rule?
I'm helping a friend out on his Calculus homework and I'm dumbfounded on this problem. I know that the correct procedure to solve this so far is:

Y= Tan(x)^x
lnY=xlnTan(x)

Then 1/tan(x)? Which would be 1/tan(0) and thus indefinite.
Help? :3 (thanks in advance)

Here is the original question:

How to Find Lim x -> 0 of Tan(x)^x with the L'Hopital Rule? - Yahoo! Answers

I have posted a link to this topic so the OP can find my response.

We are given a limit to evaluate, so let's assume it exists, and write:

$\displaystyle \lim_{x\to0}\tan^x(x)=L$

Take the natural logarithm of both sides:

$\displaystyle \ln\left(\lim_{x\to0}\tan^x(x) \right)=\ln(L)$

$\displaystyle \lim_{x\to0}\ln\left(\tan^x(x) \right)=\ln(L)$

$\displaystyle \lim_{x\to0}x\ln\left(\tan(x) \right)=\ln(L)$

$\displaystyle \lim_{x\to0}\frac{\ln\left(\cot(x) \right)}{\frac{1}{x}}=-\ln(L)$

Now we have the indeterminate form $\displaystyle \frac{\infty}{\infty}$, so application of L'Hôpital's rule yields:

$\displaystyle \lim_{x\to0}\frac{\csc(x)\sec(x)}{\frac{1}{x^2}}=-\ln(L)$

$\displaystyle \lim_{x\to0}\frac{x\sec(x)}{\frac{\sin(x)}{x}}=-\ln(L)$

Now, using the rule:

$\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim_{x\to c}f(x)}{\lim_{x\to c}g(x)}$

and the result:

$\displaystyle \lim_{x\to0}\frac{\sin(x)}{x}=1$

we have:

$\displaystyle \lim_{x\to0}x\sec(x)=-\ln(L)$

$\displaystyle 0=\ln(L)$

$\displaystyle L=1$

 

[sabrages]

a Thousand Thanks

Thank you for answering my question! Your answer is clearer than a cloudless day, brighter than the sun, and cooler than the cold waters!

Again, many thanks for making our life easier with this clear explanation.

- Michael

 

[MarkFL]

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