What is the limit of tan(x)^sin(x) as x->0+

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What is the limit of tan(x)^sin(x) as x-->0+

Homework Statement


Lim\rightarrow0^{+}((tan(x))^{sin(x)}

Homework Equations


I know that you have to raise e to this limit and you can then bring down sin(x) to get (sin(x))(ln(tan(x))), but beyond that, I am stuck.

The Attempt at a Solution


I guess above is a bit of both an attempt and relevant equations.
 
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Well, \sin(0)=0 and \ln(\tan(0))=\infty, so If you write it in the form :\frac{\ln(\tan(x))}{\frac{1}{\sin(x)}}, you have a limit of the form \frac{\infty}{\infty} and you can apply L'hospital's rule.
 


do
lim tan(x)^sin(x)=[lim sin(x)^sin(x)]/[lim cos(x)^sin(x)]
use lim x^x=1

if you want to go forward from where were, do you know l'Hopitals rule?
lim sin(x)log(tan(x))=lim log(tan(x))/csc(x)=lim [log(tan(x))]'/[csc(x)]'
 


gabbagabbahey said:
Well, \sin(0)=0 and \ln(\tan(0))=\infty, so If you write it in the form :\frac{\ln(\tan(x))}{\frac{1}{\sin(x)}}, you have a limit of the form \frac{\infty}{\infty} and you can apply L'hospital's rule.

log(tan(0+)=-infinity

Why is L'hospital's rule suggested so frequently on limit problems?
I used to mock the Hughes-Hallett Calculus book for (among other things) omiting L'hospital's rule, but now I am beginning to understand why.

Here is an extra few for practice (all limit's x->0)

lim x/x=lim x'/x'=lim 1/1=lim 1=1
lim x^x=exp(lim [log(x)]'/[1/x]')=exp(lim (1/x)/(-1/x^2))=exp(lim -x)=exp(0)=1
lim [1/x-1/x]=lim 0'/x'=lim 0/1=0
 
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Okay, so I got ((1/(tan(x)(cos^{2}(x))))/((sin(x))/(cos^{2}(x)) = (cos(x)/(sin^{2}(x)) This limit still equals infinity though, and I can' see L'hopital's rule getting me any beyond a term with infinity in it.

EDIT: Oh, so does that mean the limit DNE?
 
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iatnogpitw said:
Okay, so I got ((1/(tan(x)(cos^{2}(x))))/((sin(x))/(cos^{2}(x)) = (cos(x)/(sin^{2}(x)) This limit still equals infinity though, and I can' see L'hopital's rule getting me any beyond a term with infinity in it.

EDIT: Oh, so does that mean the limit DNE?

Errrrmm is \frac{d}{dx}\frac{1}{\sin x} REALLY \frac{\sin x}{\cos^2 x} ?:wink:
 
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