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L'Hopital's rule for solving limit

  1. Sep 15, 2005 #1
    1. lim [(1+x)^(1/x) - e ] / x
    x ->0

    2. lim [sin(2/x)+cos(1/x)]^x
    x -> inf

    Last edited: Sep 15, 2005
  2. jcsd
  3. Sep 15, 2005 #2


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    Homework should be posted in the homework forum- and you should show us what you have tried to do. Have you considered L'Hopital's rule?
  4. Sep 16, 2005 #3
    qs 1 is not 0/0 or inf/inf from
    how do consider L'Hopital's rule???
  5. Sep 16, 2005 #4


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    Do you KNOW "L'Hopital's rule"? It is specifically for limit problems where you get things like these.

    If you have [tex]lim_{x->a} \frac{f(x)}{g{x}}[/tex] where f(a)= 0 and g(a)= 0, then the limit is the same as [tex]lim_{x->a}\frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex].

    If you get things like [tex]0^0[/tex] or [tex]\infty^{\infty}[/tex] (as your second limit), you can take logarithms to reduct to the first case.
  6. Sep 17, 2005 #5


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    #1 is in form 0 / 0.
    [tex]\lim_{x \rightarrow 0} (1 + x) ^ {\frac{1}{x}} = e[/tex]
    So the numerator will tend to 0 as x approaches 0. The denominator also tends to 0. So it's 0 / 0.
    You can use L'Hopital's rule to solve for #1.
    #2 is [tex]1 ^ \infty[/tex]
    First, you can try to take logs of both sides.
    So let [tex]y = \lim_{x \rightarrow \infty} \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x[/tex]. So:
    [tex]\ln y = \ln \left\{ \lim_{x \rightarrow \infty} \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x \right\} = \lim_{x \rightarrow \infty} \ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x[/tex]
    [tex]= \lim_{x \rightarrow \infty} x \ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] = \lim_{x \rightarrow \infty} \frac{\ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right]}{\frac{1}{x}}[/tex]
    This is 0 / 0. So again, you can apply L'Hopital's rule to find the limit.
    Viet Dao,
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