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Simple question about using L'Hopital's Rule to solve limits

  1. Jul 21, 2010 #1
    I know L'Hoptal's rule can only be use when you have indeterminate forms and such, but I do not know how to make a limit a fraction sometimes. Take this limit for example:

    The limit at which X approaches infinity of (1+[tex]\frac{1}{x}[/tex])X

    I make y = The limit at which X approaches infinity of (1+[tex]\frac{1}{x}[/tex])X

    Take the Ln of both sides to get

    lny = X*ln(1+[tex]\frac{1}{x}[/tex])

    which can be rewritten as ln(1+1/x)/(1/x) but how? I get lost in the algebra and don't know how to obtain that.
  2. jcsd
  3. Jul 21, 2010 #2


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  4. Jul 21, 2010 #3


    Staff: Mentor

    Multiplying by x is the same as dividing by 1/x (assuming x is not 0).
  5. Jul 21, 2010 #4


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    One of the things you learned way back in elementary school was "to divide by a fraction, invert and multiply". Dividing by 1/x is the same inverting to get x and then multiplying:
    [itex]x ln(1+ 1/x)= ln(1+ 1/x)/(1/x)[/itex]

    This also works the other way: multiplying by a fraction is the same as dividing by its reciprocal: multiplying by x (which is the fraction x/1) is the same as dividing by its reciprocal, 1/x.
  6. Jul 21, 2010 #5
    Holy crap. Thanks guys. It's weird how I forget simple stuff but can do the "calculus" fine.
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