Indeterminate limit, using l'hopitals

**nautica17** · October 4th 2009, 04:40 PM

I have to find the following:

limit as x goes to 0 of x^sin(x)

I know that I get an indeterminate form 0^0 but what do I do from there to find the actual limit? I know the answer should be equal to 1, but how do I set up the problem to get that? Do I use l'hopitals? And if so.. how?

**Chris L T521** · October 4th 2009, 04:48 PM

Originally Posted by nautica17

I have to find the following:

limit as x goes to 0 of x^sin(x)

I know that I get an indeterminate form 0^0 but what do I do from there to find the actual limit? I know the answer should be equal to 1, but how do I set up the problem to get that? Do I use l'hopitals? And if so.. how?

Let $L=\lim_{x\to0}x^{\sin x}$ .

Then $\ln L=\lim_{x\to 0}\frac{\ln x}{\csc x}\rightarrow \frac{\infty}{\infty}$

Now apply L'Hôpital's rule.

Can you take it from here?

**artvandalay11** · October 4th 2009, 05:00 PM

Originally Posted by nautica17

I have to find the following:

limit as x goes to 0 of x^sin(x)

I know that I get an indeterminate form 0^0 but what do I do from there to find the actual limit? I know the answer should be equal to 1, but how do I set up the problem to get that? Do I use l'hopitals? And if so.. how?

For questions similar to this, where the exponent plays a key role, it helps to consider an entirely different question, namely the logarithm of the limit

So let's consider $\lim_{x\rightarrow 0}\ln (x^{\sin(x)})=\lim_{x\rightarrow 0}\sin(x)\cdot \ln(x)=\lim_{x\rightarrow 0}\frac{\ln(x)}{\frac{1}{\sin(x)}}$

Now we can apply that L'hospital's rule to get

$\displaystyle \lim_{x\rightarrow 0}\frac{\frac{1}{x}}{\frac{-\cos(x)}{\sin^2(x)}}$

$\lim_{x\rightarrow 0}\frac{-\sin^2(x)}{x\cos(x)}$

And via another LH Rule

$=\lim_{x\rightarrow 0}\frac{-2\sin(x)\cos(x)}{\cos(x)-x\sin(x)}$

=0

So the ln of the limit=0 so what number do we plug into ln to get 0? The answer is 1

There's gotta be a better way, but this is what I came up with....

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