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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


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    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.
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