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Question Number 143532 by bramlexs22 last updated on 15/Jun/21

$$\underset{x\to 0}{\lim }\frac{1-\sqrt{1+x^2}\cos x}{\tan {}^{4}x}=?$$

Answered by mathmax by abdo last updated on 15/Jun/21

$$\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\mathrm{and}\sqrt{1+{\mathrm{x}}^2}\sim 1+\frac{{\mathrm{x}}^2}{2}\Rightarrow \sqrt{1+{\mathrm{x}}^2}\mathrm{cosx}\sim (1-\frac{{\mathrm{x}}^2}{2})(1+\frac{{\mathrm{x}}^2}{2})$$$$=1-\frac{{\mathrm{x}}^4}{4}\Rightarrow 1-\sqrt{1+{\mathrm{x}}^2}\mathrm{cosx}\sim \frac{{\mathrm{x}}^4}{4}\mathrm{and}{\tan }^4\mathrm{x}\sim {\mathrm{x}}^4\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{1-\sqrt{1+{\mathrm{x}}^2}\mathrm{cosx}}{{\tan }^4\mathrm{x}}=\frac{1}{4}$$

Commented by bobhans last updated on 15/Jun/21

$$ithinktheansweris\frac{1}{3}$$

Commented by Mathspace last updated on 15/Jun/21

$$showyourworksir$$

Commented by bobhans last updated on 15/Jun/21

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