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December 7th, 2016, 11:03 PM   #1
Joined: Dec 2016
From: Việt Nam

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Calculation limit

I'm trying to figure out how to find the limit of (ln x / ln(x+1) )^xlnx , as x approaches infinity . The result is 1/e .How do I approach this?
P/s : I'm not good eng, so sorry if have anything bad
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December 8th, 2016, 12:03 AM   #2
Joined: Apr 2015
From: USA

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You can use the approximation that $\ln(1+x)\approx x$ when $x$ approaches zero.

To figure out the limit, first take the log. As a placeholder for the limit, let

$\begin{align}y&=\left[\frac{\ln x}{\ln(x+1)}\right]^{x\,\ln x}\cr
\ln y&=x\,\ln x\cdot\ln\left[\frac{\ln x}{\ln(x+1)}\right]\cr
&=x\,\ln x\cdot\ln\left[\frac{\ln x}{\ln\left(x\{1+\frac{1}{x}\}\right)}\right]\cr
&=x\,\ln x\cdot\ln\left[\frac{\ln x}{\ln x+\ln\{1+\color{red}{\frac{1}{x}}\}}\right]\cr

Since $x\to\infty$ and $\color{red}{\frac{1}{x}}$ approaches zero, we can use the approximation above.

$\begin{align}\ln y&\approx x\,\ln x\cdot\ln\left[\frac{\ln x}{\ln x+\color{red}{\frac{1}{x}}}\right]\cr
&\approx x\,\ln x\cdot\ln\left[\frac{\ln x+\frac{1}{x}}{\ln x+\frac{1}{x}}+\frac{-\frac{1}{x}}{\ln x+\frac{1}{x}}\right]\cr
&\approx x\,\ln x\cdot\ln\left[1+\frac{-\frac{1}{x}}{\ln x+\frac{1}{x}}\right]\cr

The $\frac{-\frac{1}{x}}{\ln x+\frac{1}{x}}$ also approaches zero. Use the approximation above, again.

$\begin{align}\ln y&\approx x\,\ln x\cdot\left[\frac{-\frac{1}{x}}{\ln x+\frac{1}{x}}\right]\qquad\text{The addition of }\frac{1}{x}\text{ is negligible compared to }\ln x\text{, so}\cr
&\approx x\,\ln x\cdot\left[\frac{-\frac{1}{x}}{\ln x}\right]\cr
&\approx x\,\cancel{\ln x}\cdot\left[\frac{-\frac{1}{x}}{\cancel{\ln x}}\right]\cr
&\approx x\cdot\left[-\frac{1}{x}\right]\cr
&\approx -1\cr
y&\approx e^{-1}\cr
Thanks from SenatorArmstrong
limiTS is offline  
December 8th, 2016, 04:21 AM   #3
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From: Việt Nam

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That's great. thank you very much
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