limit-x-1-1-x-x-2-

Question Number 2478 by Syaka last updated on 21/Nov/15

l i m i t x \to \infty

{(1 + \frac{1}{x})}^{x + 2} = ?

Commented by John_Haha last updated on 21/Nov/15

e

Answered by Yozzi last updated on 21/Nov/15

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

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

L = ({(1 + \frac{1}{\infty})}^{2}) (e)

L = 1^{2} \times e = e

P r o o f f o r \lim_{x \to \infty} {(1 + 1 / x)}^{x} = e

L e t l = \lim_{x \to \infty} {(1 + 1 / x)}^{x} (*) .

{(1 + 1 / x)}^{x} > 0 \forall x > 0 s o i f l e x i s t s, l > 0 .

T a k i n g l o g s t o b a s e e o n b o t h s i d e s o f (*) .

l n l = \lim_{x \to \infty} l n {(1 + 1 / x)}^{x}

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

L e t u = 1 / x \Rightarrow a s x \to \infty, u \to 0 .

∴ l n l = \lim_{u \to 0} \frac{1}{u} l n (1 + u)

U s i n g t h e M a c l a u r i n e x p a n s i o n f o r

l n (1 + u) = u - \frac{u^{2}}{2} + \frac{u^{3}}{3} - \frac{u^{4}}{4} + \frac{u^{5}}{5} - \dots - 1 < u ⩽ 1

\Rightarrow l n l = \lim_{u \to 0} \frac{1}{u} (u - \frac{u^{2}}{2} + \frac{u^{3}}{3} - \frac{u^{4}}{4} + \frac{u^{5}}{5} - \dots)

l n l = \lim_{u \to 0} (1 - \frac{u}{2} + \frac{u^{2}}{3} - \frac{u^{3}}{4} + \frac{u^{4}}{5} - \dots)

l n l = 1 - 0 + 0 - 0 + 0 - \dots

l n l = 1 \Rightarrow l = e

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

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