How do I evaluate the limit given ?

$\underset{x\to \mathrm{\infty}}{lim}\frac{{x}^{x}-{(x-1)}^{x}}{{x}^{x}}$

veleumnihryz
2022-04-25
Answered

How do I evaluate the limit given ?

$\underset{x\to \mathrm{\infty}}{lim}\frac{{x}^{x}-{(x-1)}^{x}}{{x}^{x}}$

You can still ask an expert for help

kayuukor9c

Answered 2022-04-26
Author has **13** answers

Then

$\underset{x\to \mathrm{\infty}}{lim}\frac{{x}^{x}-{(x-1)}^{x}}{{x}^{x}}=\underset{x\to \mathrm{\infty}}{lim}[1-{\left(\frac{x-1}{x}\right)}^{x}]=\underset{x\to \mathrm{\infty}}{lim}[1-{(1-\frac{1}{x})}^{x}]$

Notice

$\underset{x\to \mathrm{\infty}}{lim}{(1-\frac{1}{x})}^{x}={e}^{\underset{x\to \mathrm{\infty}}{lim}x\mathrm{ln}(1-\frac{1}{x})}$

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:

$e}^{\underset{u\to 0}{lim}\frac{\mathrm{ln}(1-u)}{u}}={e}^{\underset{u\to 0}{lim}\frac{1}{u-1}$

Notice

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