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

  1. Dec 12, 2008 #1
    What is the limit of tan(x)^sin(x) as x-->0+

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


    2. Relevant equations
    I know that you have to raise e to this limit and you can then bring down sin(x) to get [tex](sin(x))(ln(tan(x)))[/tex], 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. jcsd
  3. Dec 12, 2008 #2

    gabbagabbahey

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    Re: Limits

    Well, [itex]\sin(0)=0[/itex] and [itex]\ln(\tan(0))=\infty[/itex], so If you write it in the form :[tex]\frac{\ln(\tan(x))}{\frac{1}{\sin(x)}}[/tex], you have a limit of the form [tex]\frac{\infty}{\infty}[/tex] and you can apply L'hospital's rule.
     
  4. Dec 12, 2008 #3

    lurflurf

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    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)]'
     
  5. Dec 12, 2008 #4

    lurflurf

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    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
  6. Dec 12, 2008 #5
    Re: Limits

    Okay, so I got [tex]((1/(tan(x)(cos^{2}(x))))/((sin(x))/(cos^{2}(x)) = (cos(x)/(sin^{2}(x))[/tex] 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
  7. Dec 12, 2008 #6

    gabbagabbahey

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    Re: Limits

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