Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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.

share|cite|improve this answer
@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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.