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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9"I know what I have to use". What is that, then? – naslundx Nov 02 '14 at 11:09
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https://math.stackexchange.com/questions/295584/proof-for-ez-lim-limits-x-rightarrow-infty-left-1-fraczx-rig?noredirect=1&lq=1 – Guy Fsone Nov 10 '17 at 16:19
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https://math.stackexchange.com/questions/295584/proof-for-ez-lim-limits-x-rightarrow-infty-left-1-fraczx-rig?noredirect=1&lq=1 – Guy Fsone Nov 10 '17 at 20:36
4 Answers
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Let $f(x)=\left(\frac{x}{x+1}\right)^x$. Find the limit $$ \lim_{x\rightarrow\infty}\ln(f(x)). $$
impartialmale
- 826
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$$\lim_{x\to\infty}\Bigl(\frac x{1+x}\Bigr)^x=\lim_{x\to\infty}\Bigl(\frac{x-1}x\Bigr)^{x-1}=\lim_{x\to\infty}\Bigl(1-\frac1x\Bigr)^{x-1}=\lim_{x\to\infty}\Bigl(1-\frac1x\Bigr)^x\Big/\Bigl(1-\frac1x\Bigr)=e^{-1}/\,1=1/e$$
user2345215
- 16,422
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Suince nobody wrote the expression I was thinking of, I will:
$$\frac x{x+1}=\frac1{\frac{x+1}x}=\frac1{1+\frac1x}\implies\left(\frac x{x+1}\right)^x=\frac1{\left(1+\frac1x\right)^x}\xrightarrow[x\to\infty]{}\frac1e$$
Timbuc
- 34,191
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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 $\lim_{x\to \infty}(1+\frac 1x)^x=e$
Paul
- 20,553