Limit Problem: Solving \lim_{x\rightarrow 0} (\sin(x))^x with L'Hopital's Rule

[miglo]

Homework Statement

\lim_{x\rightarrow 0} (\sin(x))^x

Homework Equations

The Attempt at a Solution

\lim_{x\rightarrow 0} (\sin(x))^x=0^0
y=(\sin(x))^x \Rightarrow \ln(y)=x\ln(\sin(x))
\lim_{x\rightarrow 0} \ln(y) \Rightarrow \lim_{x\rightarrow 0} x\ln(\sin(x))
\lim_{x\rightarrow 0} x\ln(\sin(x))=0*\infty

after this i applied L'hopital's rule but i don't seem to be getting anywhere, i know the limit turns out to be 1, but i can't figure out what i have to do to get to that answer.
any help on what my next step should be would be greatly appreciated.

 

[Fellowroot]

Are you sure that the problem is not

limx→0⁺(sin(x))^x

Meaning only a right sided limit. Instead of both sides.

The reason why I ask is because of something we will get to in a second.

Your step here is good!

limx→0ln(y)⇒limx→0xln(sin(x))

But don't take the limit of both sides but instead only take Ln( ).

ln(y)⇒limx→0⁺xln(sin(x)) This is what you want to have.

Now, see that x in front of the Ln? How can we rewrite it to put into LHopitals form?

We can say x = 1/x^-1

so now we have

ln(y)⇒limx→0ln(sin(x))/x^-1

ln(y)⇒limx→0ln(sin(x))/(1/x)

Now graph Ln(sin(x)) on your calculator.

You will see that this graph is only a 1 sided limit coming from the right. So that's why I think that was a typo in your original problem.

Notice that limx→0⁺ln(sin(x)) is -\infty

and limx→0⁺1/x is +\infty

That means -\infty/\infty Yea! Now we can use LHopitals rule!

I'll let you finish it. Just use LHopitals rule on

ln(y)⇒limx→0ln(sin(x))/(1/x)

But don't forget after you do that, you still need to solve it for y.

So change from Log form to exponent form in the very end.

Also I had a question to the people who post here.

This is my very first post, I'm brand new.

Why can we not give complete worked out solutions to the people asking question?

I really don't understand why you can't. What's the reason. Thanks.

 

[hunt_mat]

The function is only well defined for x>0 anyway.