L'Hopital Limit

**xxsteelxx** · October 24th 2009, 10:17 AM

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

**tonio** · October 24th 2009, 12:16 PM

Originally Posted by xxsteelxx

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

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

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

so that $\left(\left(1 + \frac{1}{x}\right)^x\right)'$ $=\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]$

Applying now L'Hospital to your limit we get $-x^2\left(1 + \frac{1}{x}\right)^x \left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right]$ .

Let apart the $-\left(1 +\frac{1}{x}\right)^x$ term, whose limit is $-e$ , and focus on the rest:

$\frac{\ln \left(1+\frac{1}{x}\right)-\frac{1}{x+1}}{\frac{1}{x^2}}$

Apply L'Hospital one again here and get $\frac{-\frac{1}{x(x+1)}+\frac{1}{(x+1)^2}}{-\frac{2}{x^3}}$ $=\frac{1}{2}\left[\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2}\right]$ $=\frac{1}{2}\frac{x^2}{(x+1)^2} \xrightarrow [x \to \infty] {} \frac{1}{2}$

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

Tonio

**xxsteelxx** · October 25th 2009, 09:43 PM

Thank you very much!

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