Prove-that-e-x-Limit-u-1-x-u-u-

Question Number 6944 by Tawakalitu. last updated on 03/Aug/16

P r o v e t h a t :

e^{x} = L i m i t (u \to + \infty) {(1 + \frac{x}{u})}^{u}

Answered by sou1618 last updated on 03/Aug/16

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

\lim_{p \to + \infty} {(1 + \frac{1}{p})}^{p} = e

n o w \dots p = \frac{u}{x}

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

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

L = e^{x}

Commented by Tawakalitu. last updated on 03/Aug/16

T h a n k y o u v e r y m u c h

Answered by FilupSmith last updated on 03/Aug/16

Prove e^{x} = \lim_{u \to \infty} {(1 + \frac{x}{u})}^{u}

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

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

L = \lim_{u \to \infty} \exp (\frac{\ln (1 + \frac{x}{u})}{\frac{1}{u}})

L^{'} {H o p i t a l}^{'} s L a w

L = \lim_{u \to \infty} \exp (\frac{- \frac{x}{u^{2} + u x}}{- \frac{1}{u^{2}}})

L = \lim_{u \to \infty} \exp (\frac{u x}{u + x})

A p p l y l a w a g a i n

L = \lim_{u \to \infty} \exp (\frac{x}{1})

∴ L = e^{x}

Commented by Tawakalitu. last updated on 03/Aug/16

T h a n k s s o m u c h