# Finding the limit of a trig function

1. Sep 20, 2007

### 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.

2. Sep 20, 2007

### genneth

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

3. Sep 20, 2007

### fk378

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

4. Sep 21, 2007

### Gib Z

The easiest way is genneths second comment. Just realise 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.

5. Sep 21, 2007

### 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.

6. Sep 21, 2007

### 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.