# Finding the limit of a trig function

by lLovePhysics
Tags: function, limit, trig
 P: 167 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.
 P: 981 You should use l'Hopital's rule. Or just note that 1/0 is never going to be a good thing...
 P: 368 Would the squeeze theorem work? -1
HW Helper
P: 3,353

## Finding the limit of a trig function

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.
 P: 1,017 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.
 P: 981 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.

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