title says everything. How do I evaluate the limit given ?
How to evaluate the following limit $\lim_{x\rightarrow \infty} \frac{x^x - (x-1)^x}{x^x}$?
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3Stefano, what have you tried so far? – 2011-02-10
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0@DJC : plotting it in gnuplot :) – 2011-02-10
1 Answers
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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2@yunone: $\lim_{x \Rightarrow \infty} (1 - \frac{1}{x})^x$ is the definition of $e^{-1}$ – 2011-02-10
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0@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. – 2011-02-10
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2@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 $e^{-1}$. – 2011-02-10
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0@Sivaram, Ah, I was unaware of that! Thank you, I have learned something new then. – 2011-02-10
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0ok so the final limit is $1-e^{-1}$. – 2011-02-10
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0@yunone: You might want to look up this thread http://math.stackexchange.com/questions/17711/why-do-we-need-to-prove-euv-euev/17857#17857. It has more discussions on the definition of $e^x$ and what the other equivalent definitions of $e^x$ are. – 2011-02-10
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0@Stefano, yes, that is the correct limit. @Sivaram, thanks for the link, I remember when that thread first popped up. I'll take a closer look now. – 2011-02-11
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0You know... the most interesting fact is that now I have no idea why I asked it... :) – 2011-11-22