discuss-the-lim-n-1-1-n-n-

Question Number 117865 by aurpeyz last updated on 14/Oct/20

d i s c u s s t h e {l i m}_{n \to \infty} {(1 + \frac{1}{n})}^{n}

Answered by 1549442205PVT last updated on 14/Oct/20

\underset{n \to \infty}{Lim} {(1 + \frac{1}{n})}^{n} = e \approx 2.718

e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots + \frac{x^{n}}{n!} + \dots

Hence when x = 1 we get

e = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots . + \frac{1}{n!} + \dots

Answered by mhmoud last updated on 14/Oct/20

1

Answered by Lordose last updated on 14/Oct/20

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

\ln Λ = \lim_{x \to \infty} (nln {1 + \frac{1}{n}})

set u = \frac{1}{n} \Rightarrow {n \to \infty, u \to 0}

\ln Λ = \lim_{u \to 0} (\frac{\ln (1 + u)}{u})

L {hopital}^{'} s

\ln Λ = \lim_{u \to 0} (\frac{1}{1 + u})

\ln Λ = 1

Λ = e^{1} = e

Facebook Tweet Pin