# How do I evaluate the limit given ? \lim_{x\to\infty}\frac{x^x-(x-1)^x}{x^x}

How do I evaluate the limit given ?
$\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{x}-{\left(x-1\right)}^{x}}{{x}^{x}}$
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kayuukor9c
Then
$\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{x}-{\left(x-1\right)}^{x}}{{x}^{x}}=\underset{x\to \mathrm{\infty }}{lim}\left[1-{\left(\frac{x-1}{x}\right)}^{x}\right]=\underset{x\to \mathrm{\infty }}{lim}\left[1-{\left(1-\frac{1}{x}\right)}^{x}\right]$
Notice
$\underset{x\to \mathrm{\infty }}{lim}{\left(1-\frac{1}{x}\right)}^{x}={e}^{\underset{x\to \mathrm{\infty }}{lim}x\mathrm{ln}\left(1-\frac{1}{x}\right)}$
Try making a substitution like $u=\frac{1}{x}$ 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:
${e}^{\underset{u\to 0}{lim}\frac{\mathrm{ln}\left(1-u\right)}{u}}={e}^{\underset{u\to 0}{lim}\frac{1}{u-1}}$