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Question Number 167964 by mnjuly1970 last updated on 30/Mar/22

Commented by greogoury55 last updated on 30/Mar/22

$$=\underset{x\to 0}{\lim }\frac{1-\cos \left(\frac{\sqrt{x}-\sqrt{\sin x}}{\sqrt{x}}\right)}{x^4}$$$$=\underset{x\to 0}{\lim }\frac{{2\sin }^2\left(\frac{\sqrt{x}-\sqrt{\sin x}}{2\sqrt{x}}\right)}{x^4}$$$$=\frac{1}{2}\underset{x\to 0}{\lim }{\left(\frac{\sqrt{x}-\sqrt{\sin x}}{x^2\sqrt{x}}\right)}^2$$$$=\frac{1}{2}{[\underset{x\to 0}{\lim }\frac{x-\sin x}{x^3}.\underset{x\to 0}{\lim }\frac{\sqrt{x}}{\sqrt{x}+\sqrt{\sin x}}]}^2$$$$=\frac{1}{2}{[\frac{1}{6}.\underset{x\to 0}{\lim }\frac{1}{1+\sqrt{\frac{\sin x}{x}}}]}^2$$$$=\frac{1}{2}\times \frac{1}{144}=\frac{1}{288}$$

Answered by qaz last updated on 30/Mar/22

$$\sqrt{\frac{\sin \mathrm{x}}{\mathrm{x}}}={(1-\frac{1}{6}{\mathrm{x}}^2+\frac{1}{120}{\mathrm{x}}^4+...)}^{1/2}=1+\frac{1}{2}(-\frac{1}{6}{\mathrm{x}}^2+...)+...=1-\frac{1}{12}{\mathrm{x}}^2+...$$$$\cos (1-\sqrt{\frac{\sin \mathrm{x}}{\mathrm{x}}})=1-\frac{1}{2}{(1-\sqrt{\frac{\sin \mathrm{x}}{\mathrm{x}}})}^2+...=1-\frac{1}{2}{(\frac{1}{12}{\mathrm{x}}^2+...)}^2+...=1-\frac{1}{288}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right)$$$$\underset{\mathrm{x}\to 0}{\lim }\frac{1-\cos (1-\sqrt{\frac{\sin \mathrm{x}}{\mathrm{x}}})}{{\mathrm{x}}^4}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{288}{\mathrm{x}}^4+\mathrm{o}\left({\mathrm{x}}^4\right)}{{\mathrm{x}}^4}$$$$=\frac{1}{288}$$

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