calculate-lim-x-0-ln-1-x-x-x-2-

Question Number 32991 by abdo imad last updated on 09/Apr/18

c a l c u l a t e {l i m}_{x \to 0} \frac{l n (1 + x) - x}{x^{2}} .

Commented by prof Abdo imad last updated on 09/Apr/18

f o r ∣ x ∣< 1 {(l n (1 + x))}^{^{'}} = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}

\Rightarrow l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n}

= x - \frac{x^{2}}{2} + o (x^{3}) \Rightarrow l n (1 + x) - x = - \frac{x^{2}}{2} + o (x^{3})

\Rightarrow \frac{l n (1 + x) - x}{x^{2}} = - \frac{1}{2} + o (x) \Rightarrow

{l i m}_{x \to 0} \frac{l n (1 + x) - x}{x^{2}} = \frac{- 1}{2} .

Answered by kyle_TW last updated on 09/Apr/18

L^{'} H {\overset{}{o p i t a l}}^{'} s r u l e

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

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

= \underset{x \to 0}{l i m} (\frac{\frac{1}{1 + x} - 1}{2 x})

= \underset{x \to 0}{l i m} (\frac{1 - 1 + x}{2 x (1 + x)})

= \underset{x \to 0}{l i m} (\frac{1}{2 (1 + x)}) = \frac{1}{2}

Commented by MJS last updated on 09/Apr/18

you made one small mistake

\frac{\frac{1}{1 + x} - 1}{2 x} = \frac{1 - 1 - x}{2 x (1 + x)}

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