1. L'Hopital Limit

$\displaystyle \lim_{x\to\infty}x{[(1+\frac{1}{x})^x -e]}$ I know its indeterminate form $\displaystyle \infty*0$ and so i rewrite it as
$\displaystyle \lim_{x\to\infty}\frac{(1+\frac{1}{x})^x -e}{\frac{1}{x}}$. Now it's $\displaystyle \frac{0}{0}$ indeterminate form. Then I get stuck.

2. Originally Posted by xxsteelxx
$\displaystyle \lim_{x\to\infty}x{[(1+\frac{1}{x})^x -e]}$ I know its indeterminate form $\displaystyle \infty*0$ and so i rewrite it as
$\displaystyle \lim_{x\to\infty}\frac{(1+\frac{1}{x})^x -e}{\frac{1}{x}}$. Now it's $\displaystyle \frac{0}{0}$ indeterminate form. Then I get stuck.

This is really a tough limit: first, and in order to derivate more easily, put

$\displaystyle \displaystyle {\left(1 + \frac{1}{x}\right)^x= e^{x\ln \left(1 + \frac{1}{x}\right)}}$

so that $\displaystyle \left(\left(1 + \frac{1}{x}\right)^x\right)'$$\displaystyle =\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right] Applying now L'Hospital to your limit we get \displaystyle -x^2\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]. Let apart the \displaystyle -\left(1 +\frac{1}{x}\right)^x term, whose limit is \displaystyle -e , and focus on the rest: \displaystyle \frac{\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}}{\frac{1}{x^2}} Apply L'Hospital one again here and get \displaystyle \frac{-\frac{1}{x(x+1)}+\frac{1}{(x+1)^2}}{-\frac{2}{x^3}} \displaystyle =\frac{1}{2}\left[\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2}\right]$$\displaystyle =\frac{1}{2}\frac{x^2}{(x+1)^2} \xrightarrow [x \to \infty] {} \frac{1}{2}$

Now don't forget the term you left out before and the limit finally is (**Pant, pant!!**) : $\displaystyle -\frac{e}{2}$

Tonio

3. Thank you very much!