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Hello I am trying to find the limit of

$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $

I've tried applying L'H rule but it ends up getting really messy.

The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I can apply the standard limit laws.

Pyrons
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  • you can try to take log and proceed , makes thing simpler – avz2611 Feb 05 '15 at 11:36
  • Forget L'Hopital: $$\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\Bigg(1+\frac{1}{x}\Bigg)^{x}\Bigg(1+\frac{2}{x}\Bigg)^{-x}\Bigg(1+\frac{1}{x}\Bigg)\Bigg(1+\frac{2}{x}\Bigg)^{-2}\to e\cdot e^{-2}\cdot 1\cdot 1$$ – Did Feb 05 '15 at 11:40

6 Answers6

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$$\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\frac x{x+2}\cdot\left(1-\frac1{x+2}\right)^{x+2}\left(1-\frac1{x+2}\right)^{-1}\xrightarrow[x\to\infty]{}1\cdot\frac1e\cdot1=\frac1e$$

Timbuc
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$$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\dfrac{(x+1)^{x+2}-(x+1)^{x+1}}{(x+2)^{x+2}}$$ $$\dfrac {x(x+1)^{x+1}}{(x+2)^{x+2}}=\Big(1-\dfrac{1}{x+2}\Big)^{x+2}\Big(\dfrac{x+2}{x+1}\Big)-\Big(1-\dfrac{1}{x+2}\Big)^{x+1}\dfrac{1}{x+2}\to e^{-1}+0$$

Bumblebee
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[\begin{gathered} \mathop {\lim }\limits_{x \to \infty } \frac{{x{{\left( {x + 1} \right)}^{x + 1}}}} {{{{\left( {x + 2} \right)}^{x + 2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{x} {{x + 2}}.\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 1}} {{x + 2}}} \right)^{x + 1}} = 1.\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{1} {{x + 2}}} \right)^{x + 1}} \hfill \\ = \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 - \frac{1} {{x + 2}}} \right)}^{ - \left( {x + 2} \right)}}} \right]^{ - \frac{{x + 1}} {{x + 2}}}} = {\left[ {\mathop {\lim }\limits_{x \to \infty } {{\left( {1 - \frac{1} {{x + 2}}} \right)}^{ - \left( {x + 2} \right)}}} \right]^{\mathop {\lim }\limits_{x \to \infty } - \frac{{x + 1}} {{x + 2}}}} = {e^{ - 1}} = \frac{1} {e} \hfill \\ \end{gathered} ]

Baily
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Start with$$\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\frac{x}{x+2}\cdot\left(\frac{x+1}{x+2}\right)^{x+1},$$then note that$$\left(\frac{x+2}{x+1}\right)^{x+1}=\left(1+\frac{1}{x+1}\right)^{x+1}\longrightarrow e.$$

Amitai Yuval
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$$\lim_{x\rightarrow \infty }\frac{x(x+1)^{x+1}}{(x+2)^{x+2}}=\\\lim_{x\rightarrow \infty }\frac{(x+1)^{x+1}}{(x+2)^{x+1}}*\frac{x}{x+2}=\\\lim_{x\rightarrow \infty }(\frac{x+1}{x+2})^{x+1}\frac{x}{x+2}=\\(1-\frac{1}{x+2})^{x+1}* \lim_{x\rightarrow \infty }\frac{x}{x+2}=\\e^{-1}*\lim_{x\rightarrow \infty }\frac{x}{x+2}\\e^{-1}*1\\=\frac{1}{e}$$

Khosrotash
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let $y = \dfrac{x(x+1)^{x+1}}{(x+2)^{x+2}},$ then

$\begin{align} \ln y &= \ln x + (x+1)\ln(x+1)-(x+2)\ln(x+2)\\ &=\ln x +(x+1)[\ln x + \ln(1 + 1/x)] -(x+2)[\ln x+ \ln(1 + 2/x)]\\ &=\left(1+x+1-x-2 \right)\ln x +(x+1)\left(1/x + \cdots\right)-(x+2)\left(2/x +\cdots\right)\\ &=-1 + \cdots \end{align}$

so $$\lim_{x \to \infty} \ln y= -1 \text{ and } \lim_{x \to \infty} y= \frac{1}{e}.$$

abel
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