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Trick complicated limit

  1. Dec 21, 2011 #1
    How do I find the following limit?

    [itex]lim_{x\rightarrow 0} (\frac{\tan x}{x})^{1/x^2}[/itex]

    L'Hospital's rule gets really messy, and I cannot find a suitable
    choice of functions to apply the squeeze theorem. Does anyone
    have any suggestions? Thank you.
     
  2. jcsd
  3. Dec 21, 2011 #2

    dextercioby

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    Science Advisor
    Homework Helper

    Compute the natural logarithm of the limit with the help of l'Ho^pital's rule.
     
  4. Dec 21, 2011 #3
    Thank you. That makes life so much easier. I forgot about that little trick.
    I got the answer.
     
  5. Dec 21, 2011 #4

    Mark44

    Staff: Mentor

    It shouldn't, because L'Hopital's Rule doesn't apply here. It applies to quotients of functions -- f(x)/g(x) -- where both functions are approaching zero or both are approaching infinity.

    And you can't simply use L'Hopital's Rule on the part being raised to the power, because the limit variable occurs in the exponent.
     
  6. Dec 21, 2011 #5
    That's true, so we have

    [itex]
    \stackrel{lim}{x \to 0} \frac{\tan x}{x}^{1/x^2}= e^{\stackrel{lim}{x \to 0}\frac{ln \frac{\tan x}{x}}{\frac{1}{x^2} }}
    [/itex]

    Now,

    [itex]
    \stackrel{lim}{x \to 0} \frac{\tan x}{x}=1
    [/itex]

    hence,
    [itex]
    ln 1=0
    [/itex]

    and

    [itex]
    \stackrel{lim}{x \to 0} \frac{1}{x^2}=0
    [/itex]

    Therefore we can use l'Hospital's Rule here with [itex]f=ln \frac{\tan x}{x} [/itex] and [itex]g=1/x^2 [/itex]
     
  7. Dec 21, 2011 #6

    Mark44

    Staff: Mentor

    What you wrote is pretty hard to read and comprehend, so I can't tell if you're on the right track, plus some of what you wrote is just wrong.

    [tex]\lim_{x \to 0}\frac{1}{x^2} \neq 0![/tex]

    When you work this problem, don't include the limit until you're ready to take the limit.

    Write y = (tan(x)/x)^(1/x^2)
    Now take ln of both sides, and you'll have a quotient.
    Take the limit of ln y and of what you have on the right, and use L'Hopital's Rule on that. This is what dextercioby was saying.

    Keep in mind that the limit you'll get is the limit of the ln of something, so you'll need to adjust for that.

    LaTeX tips
    1) For complicated fractions, use [ tex ] tags instead of [ itex ] tags.
    2) For limits, don't use \stackrel -- use \lim, like this:
    [ tex ]\lim_{x \to {\infty}\frac{f(x)}{g(x)}[ /tex ]
     
  8. Dec 21, 2011 #7
    Thank you very much for the LaTeX tips. Sorry, I'm new to PF and was not aware that tex can be used. Oh I see, the i in itex stands for inline instead of the usual $. That explains a lot. Also, I made a mistake when quickly typing this up, so yes, there is a mistake.

    The correct response should be

    [tex]\lim_{x \to {\infty}} \left ( \frac{\tan x}{x} \right ) ^{1/x^2}=e^{\lim_{x \to {\infty}} \frac{1}{x^2} ln \frac{\tan x}{x} } [/tex]

    So,

    [tex] f(x)=ln \frac{\tan x}{x} [/tex]
    [tex] g(x)=x^2 [/tex]

    Now [itex] f(x) \to 0 [/itex] and [itex] g(x) \to 0 [/itex]
     
  9. Dec 21, 2011 #8

    Dick

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    Science Advisor
    Homework Helper

    Ok, so now you have a 0/0 limit. What do you get for the limit using l'Hopital? This whole thing is probably worked out easier using taylor series, if you know that approach.
     
    Last edited: Dec 21, 2011
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