find-the-valur-lim-0-x-sinx-x-3-

Question Number 195840 by NANIGOPAL last updated on 11/Aug/23

f i n d t h e v a l u r

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

Commented by kokeb last updated on 11/Aug/23

f i n d t h e v a l u r

l i \underset{\to 0}{m} \frac{x - s i n x}{x^{3}} = l i \underset{\to 0}{m} \frac{1 - c o s x}{2 x^{2}}

= l i \underset{\to 0}{m} \frac{s i n x}{4 x} = \frac{1}{4} l i \underset{\to 0}{m} \frac{s i n x}{x}

= \frac{1}{4}

Commented by jabarsing last updated on 11/Aug/23

\lim_{x \to 0} \frac{x - \sin x}{x^{3}} \overset{L H}{=} \lim_{x \to 0} \frac{1 - \cos x}{3 x^{2}} \overset{L H}{=} \lim_{x \to 0} \frac{\sin x}{6 x} =

\overset{L H}{=} \lim_{x \to 0} \frac{\cos x}{6} = \frac{1}{6}

Commented by MM42 last updated on 11/Aug/23

\frac{1}{6}

Commented by jabarsing last updated on 11/Aug/23

o r : \lim_{x \to 0} \frac{x - s i n x}{x^{3}} \overset{M . S}{=} \lim_{x \to 0} \frac{x - (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \dots)}{x^{3}} =

= \lim_{x \to 0} \frac{x - x + \frac{x^{3}}{3!} - \frac{x^{5}}{5!} + \frac{x^{7}}{7!} - \dots}{x^{3}} = \frac{1}{3!} - 0 + 0 - 0 + \dots = \frac{1}{6}