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title says everything. How do I evaluate the limit given ?

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    @DJC : plotting it in gnuplot :)2011-02-10

1 Answers 1

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Perhaps try dividing? Then $ \lim_{x\to\infty}\frac{x^x-(x-1)^x}{x^x}=\lim_{x\to\infty}\left[1-\left(\frac{x-1}{x}\right)^x\right]=\lim_{x\to\infty}\left[1-\left(1-\frac{1}{x}\right)^x\right]. $

Notice $ \lim_{x\to\infty}\left(1-\frac{1}{x}\right)^x=e^{\lim_{x\to\infty}x\ln(1-\frac{1}{x})}.(*) $

Try making a substitution like $u=1/x$ to get a situation in which l'Hôpital's rule applies to find this limit. Remember this will also change the value the limit approaches. It should look something like this: $ e^{\lim_{u\to 0}\frac{\ln(1-u)}{u}}=e^{\lim_{u\to 0}\frac{1}{u-1}}. $ Apologies for the poor legibility, I hope it at least gets you started.

*Edit: As Sivaram kindly pointed out, you could use the fact that $\lim_{x\to\infty}(1-\frac{1}{x})^x=e^{-1}$ to get the result right off the bat.

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    You k$n$ow... the most i$n$teresting fact is that now I have no idea why I asked it... :)2011-11-22