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Thread: [SOLVED] stuck on a limit and out of ideas

  1. #16
    Member sinewave85's Avatar
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    Quote Originally Posted by Moo View Post
    I'm sorry, I guess I was very tired yesterday, to write such a thing.

    But I don't agree with Prove_it's method, since it's not exactly x^x here... but it looks like it gives the correct result


    I applied twice (or three times) l'Hopital's rule :

    Let's start like you did
    $\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\ln(e^x-x-1)}{1/x}$
    this is in the form $\displaystyle \frac{\infty}{\infty}$, because :
    o $\displaystyle e^x-x-1 \to 0 \Rightarrow \ln(\dots) \to -\infty$
    o $\displaystyle 1/x \to \infty$

    So we can apply l'Hopital's rule :
    $\displaystyle \ln(L)=\lim_{x \to 0} ~ \frac{\frac{e^x-1}{e^x-x-1}}{-1/x^2}=\lim_{x \to 0} ~ -x^2 \cdot \frac{e^x-1}{e^x-x-1}$
    One last thing, if everyone is not to tired of this. Part of my frustration was that for some reason I thought that to apply lHopital's rule the equation had to either be in the form $\displaystyle \frac{+\infty}{+\infty}$ or $\displaystyle \frac{-\infty}{-\infty}$. That was part of what was geting me off track. So is that not true?
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  2. #17
    Moo
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    Quote Originally Posted by sinewave85 View Post
    One last thing, if everyone is not to tired of this. Part of my frustration was that for some reason I thought that to apply lHopital's rule the equation had to either be in the form $\displaystyle \frac{+\infty}{+\infty}$ or $\displaystyle \frac{-\infty}{-\infty}$. That was part of what was geting me off track. So is that not true?
    YOu can also apply the rule when you have $\displaystyle \frac 00$ (and by the way, the signs are not important, you can also apply this when you have $\displaystyle \frac{\infty}{-\infty}$)
    and going from $\displaystyle \frac \infty \infty$ to $\displaystyle \frac 00$ is simple, you'll see why

    assume that x goes to 0 and y goes to 0 (but different from 0 !)
    $\displaystyle \frac xy$ is in the form $\displaystyle \frac 00$

    now you can divide both numerator and denominator by xy :
    $\displaystyle \frac xy=\frac{\frac{x}{xy}}{\frac{y}{xy}}=\frac{\frac 1y}{\frac 1x}$

    and this is in the form $\displaystyle \frac \infty \infty$, since x and y go to 0 !
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  3. #18
    Member sinewave85's Avatar
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    Quote Originally Posted by Moo View Post
    YOu can also apply the rule when you have $\displaystyle \frac 00$ (and by the way, the signs are not important, you can also apply this when you have $\displaystyle \frac{\infty}{-\infty}$)
    and going from $\displaystyle \frac \infty \infty$ to $\displaystyle \frac 00$ is simple, you'll see why

    assume that x goes to 0 and y goes to 0 (but different from 0 !)
    $\displaystyle \frac xy$ is in the form $\displaystyle \frac 00$

    now you can divide both numerator and denominator by xy :
    $\displaystyle \frac xy=\frac{\frac{x}{xy}}{\frac{y}{xy}}=\frac{\frac 1y}{\frac 1x}$

    and this is in the form $\displaystyle \frac \infty \infty$, since x and y go to 0 !
    I did know that l'Hopital's rule worked for the form $\displaystyle \frac{0}{0}$ (probably should have mentioned that) but I did not know that you could mix the signs on $\displaystyle \frac \infty \infty$. A valuable piece of information, that. Thanks, Moo.
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