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

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3  
Stefano, what have you tried so far? –  JavaMan Feb 10 '11 at 22:59
    
@DJC : plotting it in gnuplot :) –  Stefano Borini Feb 10 '11 at 23:19

1 Answer 1

up vote 10 down vote accepted

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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@yunone: $\lim_{x \Rightarrow \infty} (1 - \frac{1}{x})^x$ is the definition of $e^{-1}$ –  user17762 Feb 10 '11 at 23:13
    
@Siviram, thanks for that, I wasn't sure how fair it is to use that. I wanted to show how to find that based on the definition that $\lim_{x\to\infty}(1+\frac{1}{x})^x$ is the definition of $e$, but I will edit this fact in as it makes it much clearer. –  yunone Feb 10 '11 at 23:14
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@yunone: One of the definitions of $e^x$ is $\lim_{n \rightarrow \infty} (1 + \frac{x}{n})^n$. So you can use that $\lim_{x \rightarrow \infty} (1 + \frac{-1}{x})^x$ is $e^{-1}$. –  user17762 Feb 10 '11 at 23:18
    
@Sivaram, Ah, I was unaware of that! Thank you, I have learned something new then. –  yunone Feb 10 '11 at 23:20
    
ok so the final limit is $1-e^{-1}$. –  Stefano Borini Feb 10 '11 at 23:24

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