Finding the limit of a trig function

[lLovePhysics]

I'm stuck on this limit function and I don't know what to do next. Please help me out. Thanks!

\lim_{x\rightarrow 0^{+}}(\frac{\csc2x}{x})

I just turned csc2x into 1\sin2x so then I have:

\lim_{x\rightarrow 0^{+}} (\frac{1}{x\sin2x})

Then I used the trig identity: sin2x=2sinxcosx

...but I'm not sure if this is going to take me anywhere.

I know that \lim_{x\rightarrow c} \sin x = \sin c but I'm not sure how to incoporate that identity in this problem.

 

[genneth]

You should use l'Hopital's rule. Or just note that 1/0 is never going to be a good thing...

 

[fk378]

Would the squeeze theorem work? -1<sinxcosx<1 to prove that the limit = 0?

 

[Gib Z]

The easiest way is genneths second comment. Just realize that the denominator, x sin 2x, is going to 0, and there is nothing on the numerator to cancel out with. Denominator approaches 0, whole thing approach infinity.

 

[chaoseverlasting]

You can t use L'Hospitals rule. It has to be of the form 0/0 or inf/inf. Since this is neither, try to use the identity lim x->0 sinx/x =1.

 

[genneth]

To be honest, this is one of those limits which doesn't need a limit. You just put zero in, and go, "oh, it's 1/0". No funky limit taking will change that.