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Question Number 111914 by bemath last updated on 05/Sep/20

$$\underset{x\to 0}{\lim }\frac{8x^3}{\sqrt{4+\sin 6x}-\sqrt{\tan 6x+4}}?$$

Answered by bobhans last updated on 05/Sep/20

$$\underset{x\to 0}{\lim }\frac{{8\mathrm{x}}^3}{\sqrt{4+\sin 6\mathrm{x}}-\sqrt{4+\tan 6\mathrm{x}}}=$$$$\underset{x\to 0}{\lim }\frac{{8\mathrm{x}}^3}{2(\sqrt{1+\frac{\sin 6\mathrm{x}}{4}}-\sqrt{1+\frac{\tan 6\mathrm{x}}{4}})}=$$$$\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{8\mathrm{x}}^3}{(1+\frac{\sin 6\mathrm{x}}{8})-(1+\frac{\tan 6x}{8})}=$$$$\frac{1}{2}.64.\underset{x\to 0}{\lim }\frac{x^3}{\tan 6x(\cos 6x-1)}=$$$$32.\frac{1}{6}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2}{(-2\sin {}^{2}3\mathrm{x})}=-\frac{32}{6}.\frac{1}{18}$$$$=-\frac{8}{27}.$$

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