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

$$\underset{x\to 0}{\lim }\frac{x-sinx}{x^3}=?$$$$solvewithouthopitalandanyseries.$$

Answered by mehdee42 last updated on 17/Apr/23

$$ifx\to 0\Rightarrow x-sinx\sim \frac{1}{6}{x}^3$$

Answered by mathlove last updated on 18/Apr/23

$$l=\underset{x\to 0}{\lim }\frac{x-sinx}{x^3}$$$$\underset{x\to 0}{\lim }\frac{x-2sin\frac{x}{2}cos\frac{x}{2}}{x^3}=2\underset{x\to 0}{\lim }\frac{\frac{x}{2}-sin\frac{x}{2}cos\frac{x}{2}-sin\frac{x}{2}+sin\frac{x}{2}}{x^3}$$$$2\underset{x\to 0}{\lim }\frac{\frac{x}{2}-sin\frac{x}{2}}{x^3}+2\underset{x\to 0}{\lim }\frac{sin\frac{x}{2}-sin\frac{x}{2}cos\frac{x}{2}}{x^3}$$$$2\underset{x\to 0}{\lim }\frac{\frac{x}{2}-sin\frac{x}{2}}{8{\left(\frac{x}{2}\right)}^3}+2\underset{x\to 0}{\lim }\frac{sin\frac{x}{2}(1-cos\frac{x}{2})}{8\left(\frac{x}{2}\right){\left(\frac{x}{2}\right)}^2}$$$$\frac{1}{4}\underset{x\to 0}{\lim }\frac{\frac{x}{2}-sin\frac{x}{2}}{{\left(\frac{x}{2}\right)}^3}+\frac{1}{4}\underset{x\to 0}{\lim }\frac{sin\frac{x}{2}}{\frac{x}{2}}\underset{x\to 0}{\lim }\frac{1-cos\frac{x}{2}}{{\left(\frac{x}{2}\right)}^2}$$$$\frac{1}{4}l+\frac{1}{4}\times \frac{1}{2}\Rightarrow l-\frac{1}{4}l=\frac{1}{8}$$$$l=\frac{1}{6}$$$$\underset{x\to 0}{\lim }\frac{x-sinx}{x^3}=\frac{1}{6}$$

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