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Question Number 106447 by Muhsang S L last updated on 05/Aug/20

$$\underset{x\to 0}{\lim }\frac{2x+\tan 4x}{\sqrt{1-\cos 4x\cos 6x}}=?$$

Answered by john santu last updated on 05/Aug/20

$$\underset{x\to 0}{\lim }\frac{2\mathrm{x}+4\mathrm{x}-\frac{{64\mathrm{x}}^3}{3}}{\sqrt{1-(1-{2\mathrm{x}}^2)(1-{18\mathrm{x}}^2)}}=$$$$\underset{x\to 0}{\lim }\frac{6\mathrm{x}-\frac{{64\mathrm{x}}^3}{3}}{\sqrt{1-(1-{20\mathrm{x}}^2+{36\mathrm{x}}^4)}}=$$$$\underset{x\to 0}{\lim }\frac{\frac{1}{3}\mathrm{x}(18-{64\mathrm{x}}^2)}{\sqrt{{20\mathrm{x}}^2-{36\mathrm{x}}^4}}=$$$$\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{3}\mathrm{x}(18-{64\mathrm{x}}^2)}{2\mid \mathrm{x}\mid \sqrt{5-{9\mathrm{x}}^2}}=\frac{3\sqrt{5}}{5}.$$$$\mathrm{but}\underset{x\to 0^{-}}{\lim }\frac{\frac{1}{3}\mathrm{x}(18-{64\mathrm{x}}^2)}{2\mid \mathrm{x}\mid \sqrt{5-{9\mathrm{x}}^2}}=-\frac{3\sqrt{5}}{5}$$$$\mathrm{so}\mathrm{limit}\mathrm{DNE}.$$

Commented by john santu last updated on 05/Aug/20

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