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Can someone help me calculate: $$ \lim_{x \to \infty} \frac {xe^{x/2}}{1+e^x}\quad?$$

Using l'Hospital doesn't help, but I can't figure out how to do it with Taylor polynomial... it doesn't give me anything!

Help anyone?


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5 Answers 5

We have $$ \frac{xe^{x/2}}{1+e^x}=\frac{x}{e^{x/2}}\frac{1}{1+e^{-x}}. $$ So the limit when $x\rightarrow+\infty$ is $0\cdot 1=0$.

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wow ! nice ! thanks a lot ! –  theMissingIngredient Jan 1 '13 at 21:20

It should not be hard with L'Hospital's Rule, particularly if you first divide top and bottom by $e^{x/2}$.

Remark: You might instead note that the denominator is $\gt e^x$. So our function, for positive $x$, is positive and $\lt \dfrac{xe^{x/2}}{e^x}$. So our function is $\lt \dfrac{x}{e^{x/2}}$. It is not hard to see how this behaves for large $x$.

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First: $\;$ Divide the numerator and denominator by $\,e^{x/2}\,$.

Doing so gives you:

$$ \lim_{x \to \infty}\; \frac {x\,e^{x/2}}{1+e^x}\;=\; \lim _{x\to \infty}\;\left[\left(\frac{x}{e^{x/2}}\right)\left(\frac{1}{1+e^{-x}}\right)\right]\; = \; (0)\cdot(1) = 0$$ $$ $$

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If you use L'Hospital you get $$\lim_{x\to\infty} {xe^{x/2}\over 1 + e^x} = \lim_{x\to\infty}{e^{x/2} + (1/2)xe^{x/2}\over e^x}. $$ Break this into two limits and you will get it.

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It's 0. You have a larger exponential on the bottom than the top and they outweigh everything else.

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