give-the-developpement-at-integr-series-for-f-x-ln-1-x-ln-1-x-x-2-find-lim-x-0-f-x-

Question Number 29846 by abdo imad last updated on 12/Feb/18

g i v e t h e d e v e l o p p e m e n t a t i n t e g r s e r i e s f o r

f (x) = \frac{l n (1 + x) - l n (1 - x)}{x}

2) f i n d {l i m}_{x \to 0} f (x) .

Commented by maxmathsup by imad last updated on 09/Apr/19

1) w e h a v e \frac{d}{d x} (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} x^{n + 1}}{n + 1}

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} x^{n}}{n} l e t c h a n g e x b y - x \Rightarrow

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

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

= \sum_{n = 1}^{\infty} (\frac{1 - {(- 1)}^{n}}{n}) x^{n} = \sum_{n = 1}^{\infty} \frac{2}{2 n + 1} x^{2 n + 1} \Rightarrow f (x) = 2 \sum_{n = 1}^{\infty} \frac{x^{2 n}}{2 n + 1}

w i t h ∣ x ∣< 1 a n d x \neq 0

2) w e h a v e l n (1 + x) \sim x (x \in v (0)) a n d l n (1 - x) \sim - x \Rightarrow

l n (1 + x) - l n (1 - x) \sim 2 x \Rightarrow \frac{l n (1 + x) - l n (1 - x)}{x} \sim 2 \Rightarrow

{l i m}_{x \to 0} f (x) = 2 .