1-lim-x-1-1-x-x-

Question Number 3048 by Syaka last updated on 03/Dec/15

1) \underset{x \to - \infty}{l i m} {(1 + \frac{1}{x})}_{}^{x} = ?

Answered by 123456 last updated on 04/Dec/15

L = \lim_{x \to - \infty} {(1 + \frac{1}{x})}^{x}

u = \frac{1}{x}

x \to - \infty \equiv u \to 0^{-}

L = \lim_{u \to 0^{-}} {(1 + u)}^{\frac{1}{u}}

remember that

e = \lim_{x \to 0} {(1 + x)}^{\frac{1}{x}}

so

L = e

Commented by prakash jain last updated on 04/Dec/15

y = {(1 + x)}^{1 / x}

\ln y = \frac{1}{x} \ln (1 + x)

\underset{x \to 0}{\lim l n} y = \lim_{x \to 0} \frac{\ln (1 + x)}{x} = \lim_{x \to 0} \frac{\frac{d}{d x} \ln (1 + x)}{\frac{d}{d x} x} = \lim_{x \to 0} \frac{1}{1 + x} = 1

\underset{x \to 0}{\lim l n} y = 1

\lim_{x \to 0} y = e^{1} = e