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Question Number 168166 by cortano1 last updated on 05/Apr/22

$$\underset{x\to 0}{\lim }\frac{\sin \left(\sin x\right)-x\sqrt[3]{1-x}}{x^5}=?$$

Answered by qaz last updated on 05/Apr/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{\sin \left(\sin \mathrm{x}\right)-\mathrm{x}\sqrt[3]{1-\mathrm{x}}}{{\mathrm{x}}^5}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\sin (\mathrm{x}-\frac{1}{6}{\mathrm{x}}^3+...)-\mathrm{x}[1+\left(\begin{array}{c}1/3\\ 1\end{array}\right)(-\mathrm{x})+\left(\begin{array}{c}1/3\\ 2\end{array}\right){(-\mathrm{x})}^2+\left(\begin{array}{c}1/3\\ 3\end{array}\right){(-\mathrm{x})}^3+\left(\begin{array}{c}1/3\\ 4\end{array}\right){(-\mathrm{x})}^4+...]}{{\mathrm{x}}^5}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{[(\mathrm{x}-\frac{1}{6}{\mathrm{x}}^3+\frac{1}{120}{\mathrm{x}}^5+...)-\frac{1}{6}{\mathrm{x}}^3{(1-\frac{1}{6}{\mathrm{x}}^2+...)}^3+\frac{1}{120}{\mathrm{x}}^5{(1-\frac{1}{6}{\mathrm{x}}^2+...)}^5+...]-\mathrm{x}(1-\frac{1}{3}\mathrm{x}-\frac{1}{9}{\mathrm{x}}^2-\frac{5}{81}{\mathrm{x}}^3-\frac{70}{1944}{\mathrm{x}}^4+...)}{{\mathrm{x}}^5}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{3}{\mathrm{x}}^2+\mathrm{o}\left({\mathrm{x}}^2\right)}{{\mathrm{x}}^5}$$$$=\pm \mathrm{\infty }$$

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