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Question Number 159754 by cortano last updated on 20/Nov/21

$$\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\tan x}+\sqrt{\sin x}-2\sqrt{x}}{\sqrt{\sin x}-\sqrt{\tan x}}=?$$

Answered by bobhans last updated on 21/Nov/21

$$\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\tan \mathrm{x}}+\sqrt{\sin \mathrm{x}}-2\sqrt{\mathrm{x}}}{\sqrt{\sin \mathrm{x}}-\sqrt{\tan \mathrm{x}}}$$$$=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\sin \mathrm{x}}-\sqrt{\tan \mathrm{x}}+2\sqrt{\tan \mathrm{x}}-2\sqrt{\mathrm{x}}}{\sqrt{\sin \mathrm{x}}-\sqrt{\tan \mathrm{x}}}$$$$=1+2\underset{x\to 0^{+}}{\lim }\frac{\sqrt{\tan \mathrm{x}}-\sqrt{\mathrm{x}}}{\sqrt{\sin \mathrm{x}}-\sqrt{\tan \mathrm{x}}}$$$$=1+2\underset{x\to 0^{+}}{\lim }\frac{\tan \mathrm{x}-\mathrm{x}}{\sin \mathrm{x}-\tan \mathrm{x}}.\frac{\sqrt{\sin \mathrm{x}}+\sqrt{\tan \mathrm{x}}}{\sqrt{\tan \mathrm{x}}+\sqrt{\mathrm{x}}}$$$$=1+2[\underset{x\to 0^{+}}{\lim }\frac{\tan \mathrm{x}-\mathrm{x}}{{\mathrm{x}}^3}.\frac{{\mathrm{x}}^3}{\sin \mathrm{x}-\tan \mathrm{x}}.\frac{\sqrt{\frac{\sin \mathrm{x}}{\mathrm{x}}}+\sqrt{\frac{\tan \mathrm{x}}{\mathrm{x}}}}{\sqrt{\frac{\tan \mathrm{x}}{\mathrm{x}}}+1}]$$$$=1+2[\frac{1}{3}.(-2)]=1-\frac{4}{3}=-\frac{1}{3}$$

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