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Question Number 159611 by cortano last updated on 19/Nov/21

Answered by qaz last updated on 19/Nov/21

$$\underset{\mathrm{x}\to 0}{\lim }\sqrt[{\mathrm{x}}^2]{1+\sin (1-\frac{\sin \mathrm{x}}{\mathrm{x}})}$$$$=\mathrm{e}\underset{\mathrm{x}\to 0}{\lim }\frac{\ln (1+\sin (1-\frac{\sin \mathrm{x}}{\mathrm{x}}))}{{\mathrm{x}}^2}$$$$=\mathrm{e}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin (1-\frac{\sin \mathrm{x}}{\mathrm{x}})}{{\mathrm{x}}^2}$$$$=\mathrm{e}\underset{\mathrm{x}\to 0}{\lim }\frac{\sin (1-\frac{\sin \mathrm{x}}{\mathrm{x}})}{1-\frac{\sin \mathrm{x}}{\mathrm{x}}}\cdot \frac{1-\frac{\sin \mathrm{x}}{\mathrm{x}}}{{\mathrm{x}}^2}$$$$=\mathrm{e}\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3}$$$$={\mathrm{e}}^{\frac{1}{6}}$$

Answered by tounghoungko last updated on 19/Nov/21

$$\underset{x\to 0}{\lim }{(1+\sin (1-\left(\frac{\sin x}{x}\right)))}^{\frac{1}{x^2}}$$$$=e^{\underset{x\to 0}{\lim }\left(\frac{\sin \left(\frac{x-\sin x}{x}\right)}{x^2}\right)}$$$$=e^{\underset{x\to 0}{\lim }\left(\frac{\sin \left(x^2\left(\frac{x-\sin x}{x^3}\right)\right)}{x^2}\right)}$$$$=e^{\underset{x\to 0}{\lim }\left(\frac{\sin \left(\frac{1}{6}{x}^2\right)}{x^2}\right)}$$$$=e^{\frac{1}{6}}$$

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