Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit candidate for L'hopital

  1. Aug 18, 2008 #1
    [tex]\lim_{x \rightarrow infinity}/ left(\frac{\1+Tan\frac{/pi}{2x}}{1+sin\frac{/pi}{3x}}}right)^x[/tex]
  2. jcsd
  3. Aug 18, 2008 #2
    Re: limit

    sorry i made mistake in typing
  4. Aug 18, 2008 #3


    User Avatar
    Homework Helper

    Re: limit

    So what does the limit problem actually look like?
  5. Aug 18, 2008 #4
    Re: limit

    I assume you meant

    [tex] \lim_{x \rightarrow {\infty}} \left(\frac{1 \, + \, tan\left(\frac{\pi}{2x}\right)}{1 \, + \, sin \left(\frac{\pi}{3x}\right)} \right)^{x} [/tex]

    then I would suggest letting that = y, then take ln of both sides and you should get something like:

    [tex] ln(y) \, = \, \lim_{x \rightarrow {\infty}} \frac{ln\left(\frac{1 \, + \, tan\left(\frac{\pi}{2x}\right)}{1 \, + \, sin \left(\frac{\pi}{3x}\right)} \right)}{\frac{1}{x}} [/tex]

    which if you "plug in" the limit should give you [tex]\frac{0}{0}[/tex] making it a candidate for L'hopital. Try that.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Limit candidate L'hopital Date
B Question about a limit definition Feb 27, 2018
I Looking for additional material about limits and distributions Feb 17, 2018
I A problematic limit to prove Jan 26, 2018
B Proof of a limit rule Dec 19, 2017