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How can I evaluate this limit?

$$\lim\limits_{x\to\infty}\left(\dfrac x{x+1}\right)^x.$$

Hm I can't take this limit. I know what I have to use, but I can't. Sorry about this. Who can do it for example?

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closed as off-topic by Najib Idrissi, Hans Lundmark, Mathmo123, Joonas Ilmavirta, Mark Fantini Nov 2 '14 at 13:21

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$$\lim_{x\to\infty}\Bigl(\frac x{1+x}\Bigr)^x=\lim_{x\to\infty}\Bigl(\frac{x-1}x\Bigr)^{x-1}=\lim_{x\to\infty}\Bigl(1-\frac1x\Bigr)^{x-1}=\lim_{x\to\infty}\Bigl(1-\frac1x\Bigr)^x\Big/\Bigl(1-\frac1x\Bigr)=e^{-1}/\,1=1/e$$

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  • $\begingroup$ $e^{-1}/1\ne1$. Check your last step again. $\endgroup$ – Akiva Weinberger Nov 12 '14 at 11:22
  • $\begingroup$ @columbus8myhw ? $\endgroup$ – user2345215 Nov 12 '14 at 15:57
  • $\begingroup$ Never mind. My phone rendered the math weirdly. $\endgroup$ – Akiva Weinberger Nov 12 '14 at 16:01
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Let $f(x)=\left(\frac{x}{x+1}\right)^x$. Find the limit $$ \lim_{x\rightarrow\infty}\ln(f(x)). $$

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Suince nobody wrote the expression I was thinking of, I will:

$$\frac x{x+1}=\frac1{\frac{x+1}x}=\frac1{1+\frac1x}\implies\left(\frac x{x+1}\right)^x=\frac1{\left(1+\frac1x\right)^x}\xrightarrow[x\to\infty]{}\frac1e$$

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  • $\begingroup$ This is basically multiplying with $1=\frac{1/x}{1/x}$ $\endgroup$ – Alice Ryhl Nov 18 '14 at 15:40
  • $\begingroup$ Well, in fact it is exactly the same. Yes. $\endgroup$ – Timbuc Nov 18 '14 at 16:54
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Notice that $(\frac{x}{x+1})^x=(1-\frac{1}{x+1})^{x+1}\cdot \frac{1}{1-\frac{1}{x+1}}$ and use $\lim_{x\to \infty}(1+\frac 1x)^x=e$

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