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

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

$\displaystyle Let y = (1 + \frac{1}{x})^x, ln y = xln(1 + \frac{1}{x})My Solution

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

$\displaystyle lny \to 1, y \to e$

Thanks in advance.