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Limits to the power of a variable

  1. May 31, 2006 #1
    I know how to solve a regular question such as x^x, but how would I find:


    My study guide says that I have to use Ln and then L'Hopital's Rule, and I can see how that would work, but what happens to the limit?
  2. jcsd
  3. May 31, 2006 #2


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    Yes, you can use ln, and then L'Hopital's rule. However, there's a simplier way to do this.
    [tex]\lim_{x \rightarrow + \infty} \left( 1 + \frac{1}{x} \right) ^ x = \lim_{x \rightarrow - \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex]
    So ingeneral:
    [tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex]
    To solve the problem, you must try to make the expression have the form above. So, we have:
    [tex]\lim_{x \rightarrow \infty} \left( 1 - \frac{4}{x} \right) ^ x = \lim_{x \rightarrow \infty} \left( 1 - \frac{1}{\frac{x}{4}} \right) ^ x = \lim_{x \rightarrow \infty} \left( 1 + \frac{1}{\frac{- x}{4}} \right) ^ x[/tex]
    [tex]= \lim_{x \rightarrow \infty} \left[ \left( 1 + \frac{1}{\frac{- x}{4}} \right) ^ \frac{-x}{4} \right] ^ {-4}[/tex]
    Now, can you go from here? :)
    If you want to use L'Hopital's rule, you can let:
    [tex]y = \lim_{x \rightarrow \infty} \left( 1 - \frac{4}{x} \right) ^ x[/tex]
    [tex]\Rightarrow \ln y = \ln \lim_{x \rightarrow \infty} \left( 1 - \frac{4}{x} \right) ^ x = \lim_{x \rightarrow \infty} \ln \left( 1 - \frac{4}{x} \right) ^ x = \lim_{x \rightarrow \infty} \left[ x \ln \left( 1 - \frac{4}{x} \right) \right] = \lim_{x \rightarrow \infty} \left[ \frac{\ln \left( 1 - \frac{4}{x} \right) }{\frac{1}{x}} \right] = z[/tex]
    Now, you can apply L'Hopital's rule here. The RHS is the Indeterminate form 0 / 0.
    After applying L'Hopital's Rule, we can obtain y by the fact that y = ez.
    Can you go from here? :)
    Last edited: Jun 1, 2006
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