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

1. Dec 12, 2008

### iatnogpitw

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

1. The problem statement, all variables and given/known data
$$Lim\rightarrow0^{+}((tan(x))^{sin(x)}$$

2. Relevant 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.

3. The attempt at a solution
I guess above is a bit of both an attempt and relevant equations.

2. Dec 12, 2008

### gabbagabbahey

Re: Limits

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.

3. Dec 12, 2008

### lurflurf

Re: Limits

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)]'

4. Dec 12, 2008

### lurflurf

Re: Limits

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 begining 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

Last edited: Dec 12, 2008
5. Dec 12, 2008

### iatnogpitw

Re: Limits

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?

Last edited: Dec 12, 2008
6. Dec 12, 2008

### gabbagabbahey

Re: Limits

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?