lim-x-0-x-sinx-x-3-solve-without-hopital-and-any-series-

Question Number 191078 by sciencestudentW last updated on 17/Apr/23

\lim_{x \to 0} \frac{x - s i n x}{x^{3}} = ?

s o l v e w i t h o u t h o p i t a l a n d a n y s e r i e s .

Answered by mehdee42 last updated on 17/Apr/23

i f x \to 0 \Rightarrow x - s i n x \sim \frac{1}{6} x^{3}

Answered by mathlove last updated on 18/Apr/23

l = \lim_{x \to 0} \frac{x - s i n x}{x^{3}}

\lim_{x \to 0} \frac{x - 2 s i n \frac{x}{2} c o s \frac{x}{2}}{x^{3}} = 2 \lim_{x \to 0} \frac{\frac{x}{2} - s i n \frac{x}{2} c o s \frac{x}{2} - s i n \frac{x}{2} + s i n \frac{x}{2}}{x^{3}}

2 \lim_{x \to 0} \frac{\frac{x}{2} - s i n \frac{x}{2}}{x^{3}} + 2 \lim_{x \to 0} \frac{s i n \frac{x}{2} - s i n \frac{x}{2} c o s \frac{x}{2}}{x^{3}}

2 \lim_{x \to 0} \frac{\frac{x}{2} - s i n \frac{x}{2}}{8 {(\frac{x}{2})}^{3}} + 2 \lim_{x \to 0} \frac{s i n \frac{x}{2} (1 - c o s \frac{x}{2})}{8 (\frac{x}{2}) {(\frac{x}{2})}^{2}}

\frac{1}{4} \lim_{x \to 0} \frac{\frac{x}{2} - s i n \frac{x}{2}}{{(\frac{x}{2})}^{3}} + \frac{1}{4} \lim_{x \to 0} \frac{s i n \frac{x}{2}}{\frac{x}{2}} \lim_{x \to 0} \frac{1 - c o s \frac{x}{2}}{{(\frac{x}{2})}^{2}}

\frac{1}{4} l + \frac{1}{4} \times \frac{1}{2} \Rightarrow l - \frac{1}{4} l = \frac{1}{8}

l = \frac{1}{6}

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

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