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Difficult L'Hopital Problem

  1. Sep 20, 2006 #1
    I got the following problem in my math class:

    http://img81.imageshack.us/img81/7508/limitdv4.jpg [Broken]

    I know that I'm supposed to use L'Hopital's rule, but I have 2 problems. First of all, I don't know how to make that into a fraction, besides putting it all over 1 or making tan negative and putting it under 1. Also, I have no idea how to even find the derivative of that mess.

    I know that the answer is e^(2/pi), but I'd like to know how to get it.

    Can anybody help me, please?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Sep 20, 2006 #2
    Right now it is in indeterminate form, so first thing you will have to do is figure out how to fix that since L'Hopital only works for (inf/inf) and (0/0). You might try making some function i.e. L(x) equal to that limit. Then perhaps use log rules and such. I haven't actually worked it out to see if that would work since that is your job, but play around and you will get it.
  4. Sep 21, 2006 #3


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    Ok, you can solve the problem like this:
    Let: [tex]y = \lim_{x \rightarrow 1} (2 - x) ^ {\tan \left( \frac{\pi}{2} x \right)}[/tex]
    Now, by taking log of both sides, and using the fact that the function ln(x) is continuous for all x > 0, we have:
    [tex]\ln y = \ln \left( \lim_{x \rightarrow 1} (2 - x) ^ {\tan \left( \frac{\pi}{2} x \right)} \right) = \lim_{x \rightarrow 1} \left[ \ln \left( (2 - x) ^ {\tan \left( \frac{\pi}{2} x \right)} \right) \right] = \lim_{x \rightarrow 1} \left[ \tan \left( \frac{\pi}{2} x \right) \ln (2 - x) \right][/tex]
    The limit in the right is in the Indeterminate form [tex]0 \times \infty[/tex]
    Now, for it to have the form [tex]\frac{0}{0} \quad \mbox{or} \quad \frac{\infty}{\infty}[/tex], we can divide both numerator, and denominator by ln(2 - x), or tan(pi x / 2), like this:
    [tex]\ln y = \lim_{x \rightarrow 1} \left[ \tan \left( \frac{\pi}{2} x \right) \ln (2 - x) \right] = \lim_{x \rightarrow 1} \frac{\tan \left( \frac{\pi}{2} x \right)}{\frac{1}{\ln (2 - x)}}[/tex]
    Can you go from here?
    After evaluating the limit in the right, can you find y, i.e the limit of the original problem? :)
    Last edited by a moderator: May 2, 2017
  5. Sep 22, 2006 #4


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    1. Make the sub

    [tex] x=y+1 [/tex]

    2. Then

    [tex] \lim_{y\rightarrow 0^{+}} (1-y)^{\frac{1}{y}\cdot \[y\tan\left(\frac{\pi}{2}(1+y)\right)} [/tex]

    U can write it as e^(-lim...) and you may use (if you wish) l'H^opital's rule to evaluate the limit y/cos(pi/2(1+y)) as y tends to zero plus.

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