Lim x tends to infinity ( e / ( 1 + 1 / x )^x ) ^x
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Lim x tends to infinity ( e / ( 1 + 1 / x )^x ) ^x
Pls help me...
Is this $\displaystyle \displaystyle \begin{align*} \lim_{x \to \infty}\left[\frac{e}{\left(1 + \frac{1}{x}\right)^x}\right]^x \end{align*} $?Quote:
Originally Posted by Abhi
yesQuote:
Originally Posted by Prove It
Hello Friends!
Quote:
Solve $\displaystyle \lim_{x \to \infty}[\frac{e}{(1+\frac{1}{x})^x}]^x$
$\displaystyle \begin{align*} \lim_{x \to \infty}[\frac{e}{(1+\frac{1}{x})^x}]^x &= \exp \lim_{x \to \infty} x\ln[\frac{e}{(1+\frac{1}{x})^x}] \\ &=\exp \lim_{x \to \infty}x[\ln{e}-x\ln{(1+\frac{1}{x}})] \\ &=\exp\lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})] \end{align*}$
$\displaystyle \begin{align*} \lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})] &= \lim_{x \to \infty} [x-x^2\left \{ \frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-... \right \}] \\ &=\lim_{x \to \infty}[x-x+\frac{1}{2}-\frac{1}{3x}+...] \\ &=\frac{1}{2} \end{align*}$Quote:
For the limit $\displaystyle \lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})]$
Therefore: $\displaystyle \lim_{x \to \infty}[\frac{e}{(1+\frac{1}{x})^x}]^x=\exp\lim_{x \to \infty} [x-x^2\ln{(1+\frac{1}{x}})]=\exp(\frac{1}{2})=\sqrt{e}$
