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Question Number 69607 by malwaan last updated on 25/Sep/19

w i t h o u t u s i n g l h o s p i t a l p l e a s e

p r o v e t h a t

\underset{x \to 0}{l i m} \frac{x - s i n x}{x^{3}} = \frac{1}{6}

I w a n t e v e r y m e t h o d

p o s s i b l e b e c a u s e s o m e o n e

c h a l l e n g e m e

Commented by mathmax by abdo last updated on 26/Sep/19

w e h a v e s i n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1)!} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots \Rightarrow

- s i n x = - x + \frac{x^{3}}{3!} - \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + o (x^{7}) \Rightarrow x - s i n x =

\frac{x^{3}}{3!} - \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + o (x^{7}) \Rightarrow \frac{x - s i n x}{x^{3}} = \frac{1}{3!} - \frac{x^{2}}{5!} + \frac{x^{4}}{7!} + o (x^{4}) \Rightarrow

{l i m}_{x \to 0} \frac{x - s i n x}{x^{3}} = \frac{1}{3!} = \frac{1}{6}

Commented by Prithwish sen last updated on 25/Sep/19

li \underset{x \to 0}{m} \frac{x - (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots .)}{x^{3}} = li \underset{x \to 0}{m} \frac{1}{3!} - \frac{x^{2}}{5!} + the higher power of x

= \frac{1}{3!} = \frac{1}{6} proved

Commented by malwaan last updated on 26/Sep/19

t h a n k y o u s i r

Commented by malwaan last updated on 27/Sep/19

t h a n k y o u s o m u c h

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