Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Thanks!

share|improve this question

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 $0\cdot 1=0$.

share|improve this answer
    
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 $x$.

share|improve this answer


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

Doing so gives you:

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

share|improve this answer

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.