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Question Number 101366 by bobhans last updated on 02/Jul/20

$$\underset{x\to 0}{\lim }\frac{x^3-\sin {}^{3}x}{x^5}=?$$

Commented by Dwaipayan Shikari last updated on 02/Jul/20

$$\underset{x\to 0}{\lim }\frac{x^3-{(x-\frac{x^3}{6})}^3}{x^5}=\underset{x\to 0}{\lim }\frac{x^3-x^3+\frac{x^9}{6}+\frac{3x^5}{6}-\frac{3x^7}{36}}{x^5}=\frac{\frac{3x^5}{6}}{x^5}=\frac{1}{2}\left\{\frac{x^9}{6}and\frac{3x^7}{36}arenegligable\right\}$$

Commented by bemath last updated on 02/Jul/20

$$\mathrm{agree}$$

Answered by ajfour last updated on 02/Jul/20

$$\underset{x\to 0}{\lim }\frac{(x-\sin x)(x^2+x\sin x+\sin {}^{2}x)}{x^5}$$$$\underset{x\to 0}{\lim }\frac{(x-\sin x)}{x^3}\times \underset{x\to 0}{\lim }\frac{(x^2+x\sin x+\sin {}^{2}x)}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{x-(x-\frac{x^3}{3!}+....)}{x^3}\times \left(3\right)$$$$=\frac{1}{3!}\times 3=\frac{1}{2}.$$

Commented by bemath last updated on 02/Jul/20

$$\mathrm{yes}..\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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