I trying to show
$$ \lim_{x\to\infty} \left(\frac{x}{x+1} \right)^x $$
I know the answer using wolfram or a calculator, but I stuck trying to figure out how to apply L'hopital.
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Hint $$ \left(\frac{x}{x+1}\right)^x =\left(1-\frac{1}{x+1}\right)^x =\left(1-\frac{1}{u}\right)^{u-1} =\left(1-\frac{1}{u}\right)^{u}\left(1-\frac{1}{u}\right)^{-1} $$ and $x\to \infty$ iff $u\to\infty$.
Sri-Amirthan Theivendran
- 31,832
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HINT:
First note that we can write
$$\left(\frac{x}{x+1}\right)^x=e^{-x\log\left(1+\frac1x\right)}$$
Now apply L'Hospital's Rule to $\lim_{x\to \infty}\frac{\log\left(1+\frac1x\right)}{\frac1x}$
Mark Viola
- 184,670
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I understand the 1st line, but I don't understand the 2nd. Where did the 1/x on the bottom come from? What happened to the e? – Michael Maliszesky Jun 15 '17 at 11:18
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Disregard the last question I get it now. Thank you. – Michael Maliszesky Jun 15 '17 at 12:17
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I've deleted the comment. Perhaps you can do so also. – Mark Viola Jun 15 '17 at 12:28
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You're welcome. My pleasure. – Mark Viola Jun 15 '17 at 12:29
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Hint: For such $1^\infty$ problems, you can use the trick below:
$y=e^{\ln y}$
Then you can apply L'Hospital rule to the exponent part.
Huang
- 549
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As $x$ approaches infinity: $$\biggl(\frac{x}{x+1}\biggl)^x=\biggl(1-\frac{1}{x+1}\biggl)^x=e^{\biggl(-\frac{1}{x+1}\biggl)x}=e^{-1}$$ by using Euler's method.