
#1
Sep2007, 06:50 PM

P: 167

I'm stuck on this limit function and I don't know what to do next. Please help me out. Thanks!
[tex]\lim_{x\rightarrow 0^{+}}(\frac{\csc2x}{x})[/tex] I just turned csc2x into 1\sin2x so then I have: [tex] \lim_{x\rightarrow 0^{+}} (\frac{1}{x\sin2x})[/tex] Then I used the trig identity: sin2x=2sinxcosx ...but I'm not sure if this is going to take me anywhere. I know that [tex]\lim_{x\rightarrow c} \sin x = \sin c[/tex] but I'm not sure how to incoporate that identity in this problem. 



#2
Sep2007, 07:00 PM

P: 981

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




#3
Sep2007, 07:46 PM

P: 368

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




#4
Sep2107, 04:56 AM

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.




#5
Sep2107, 08:48 AM

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.




#6
Sep2107, 08:52 AM

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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