find-lim-x-0-1-x-ln-e-x-1-x-

Question Number 28892 by abdo imad last updated on 31/Jan/18

f i n d {l i m}_{x \to 0} \frac{1}{x} l n (\frac{e^{x} - 1}{x}) .

Answered by ajfour last updated on 01/Feb/18

\lim_{x \to 0} \frac{1}{x} \ln (\frac{x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + . .}{x})

= \lim_{x \to 0} \frac{\ln (1 + \frac{x}{2} + \frac{x^{2}}{6} + . .)}{x}

= \lim_{x \to 0} \frac{\frac{x}{2} + \frac{x^{2}}{6} + . .}{x} = \frac{1}{2} .

Answered by chantriachheang last updated on 01/Feb/18

L = \underset{x \to 0}{l i m} \frac{1}{x} l n (\frac{e^{x} - 1}{x}) (1)

t a k e x = - x

\Rightarrow L = \underset{x \to 0}{l i m} (- \frac{1}{x}) l n \frac{e^{x} - 1}{{x e}^{x}} (2)

(1) + (2)

\Rightarrow 2 L = \underset{x \to 0}{l i m} \frac{1}{x} [l n (\frac{e^{x} - 1}{x}) - l n (\frac{e^{x} - 1}{{x e}^{x}})]

2 L = \underset{x \to 0}{l i m} \frac{1}{x} [l n (\frac{e^{x} - 1}{x}) - l n (\frac{e^{x} - 1}{x}) - l n \frac{1}{e^{x}}]

2 L = \underset{x \to 0}{l i m} (\frac{1}{x}) (- {l n e}^{- x}) = l n e = 1

\Rightarrow L = \frac{1}{2}

:)