prove-that-lim-x-1-1-x-x-e-

Question Number 121042 by bramlexs22 last updated on 05/Nov/20

prove that \lim_{x \to \infty} {(1 + \frac{1}{x})}^{x} = e

Answered by bobhans last updated on 05/Nov/20

L e t w = \lim_{x \to \infty} {(1 + \frac{1}{x})}^{x}

t h e n \ln (w) = \ln (\lim_{x \to \infty} {(1 + \frac{1}{x})}^{x})

\ln (w) = \lim_{x \to \infty} (\ln {(1 + \frac{1}{x})}^{x})

\ln (w) = \lim_{x \to \infty} (x . \ln (1 + \frac{1}{x}))

\ln (w) = \lim_{x \to \infty} (x . [\frac{1}{x} - \frac{1}{2 x^{2}} + \frac{1}{3 x^{3}} - \frac{1}{4 x^{4}} + \dots])

\ln (w) = \lim_{x \to \infty} (1 - \frac{1}{2 x} + \frac{1}{3 x^{2}} - \frac{1}{4 x^{3}} + \dots)

\ln (w) = 1 \Rightarrow w = e^{1} = e .

Answered by Dwaipayan Shikari last updated on 05/Nov/20

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

= 1 + \frac{x}{x} + \frac{x (x - 1)}{2! x^{2}} + \frac{x (x - 1) (x - 2)}{3! x^{3}} + \dots

= 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots

= \underset{n = 0}{\sum^{\infty}} \frac{1}{n!} = e