Finding the limit of a trig function


by lLovePhysics
Tags: function, limit, trig
lLovePhysics
lLovePhysics is offline
#1
Sep20-07, 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.
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genneth
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#2
Sep20-07, 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...
fk378
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#3
Sep20-07, 07:46 PM
P: 368
Would the squeeze theorem work? -1<sinxcosx<1 to prove that the limit = 0?

Gib Z
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#4
Sep21-07, 04:56 AM
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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.
chaoseverlasting
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#5
Sep21-07, 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.
genneth
genneth is offline
#6
Sep21-07, 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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