Limit of an Indeterminate Form Problem

**mturner07** · February 25th 2010, 05:14 PM

Hey I was just wondering if I did this problem right.

$\lim_{x \to \infty}(1 + \frac{1}{x})^x$

My Solution

$Let y = (1 + \frac{1}{x})^x, ln y = xln(1 + \frac{1}{x})<br />$

$\lim_{x \to \infty} lny = \lim_{x \to \infty}\frac{ln(1 + \frac{1}{x})}{\frac{1}{x}} = \frac{0}{0},$ Using L'Hopital's Rule. $= \lim_{x \to \infty}\frac{-\frac{1}{x^2 + x}}{-\frac{1}{x^2}} = \lim_{x \to \infty}\frac{x}{x + 1} = \frac{\infty}{\infty} = \lim_{x \to \infty}\frac{1}{1} = 1$

$lny \to 1, y \to e$

Thanks in advance.

**Scott H** · February 25th 2010, 05:58 PM

Looks good to me.

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