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Question Number 107446 by mathdave last updated on 10/Aug/20

Commented by kaivan.ahmadi last updated on 10/Aug/20

$$$$$${lim}_{x\to 0}\frac{{sinx}^3}{x^4}={lim}_{x\to 0}\frac{x^3}{4x^3}=\frac{1}{4}$$

Commented by mr W last updated on 10/Aug/20

$$=\underset{x\to 0}{\lim }\frac{\sin x^3}{4x^3}=\frac{1}{4}$$

Commented by mr W last updated on 10/Aug/20

$$=\underset{x\to 0}{\lim }\frac{{\int }_0^x(t^3-\frac{t^9}{3!}+\frac{t^{15}}{5!}-...)dt}{x^4}$$$$=\underset{x\to 0}{\lim }\frac{\frac{x^4}{4}-\frac{x^{10}}{10\times 3!}+\frac{x^{16}}{16\times 5!}-...}{x^4}$$$$=\underset{x\to 0}{\lim }(\frac{1}{4}-\frac{x^6}{10\times 3!}+\frac{x^{12}}{16\times 5!}-...)$$$$=\frac{1}{4}$$

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