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Question Number 10033 by ridwan balatif last updated on 21/Jan/17

Answered by mrW1 last updated on 21/Jan/17

$$\underset{x\to 0}{\lim }\frac{x(\cos 4x^2-1)}{(1-\frac{1}{{\cos }^{2}2x})}$$$$=\underset{x\to 0}{\lim }\frac{x(\cos 4x^2-1){\cos }^{2}2x}{({\cos }^{2}2x-1)}$$$$=\underset{x\to 0}{\lim }\frac{x(-{2\sin }^{2}2x^2){\cos }^{2}2x}{-{\sin }^{2}2x}$$$$=\underset{x\to 0}{\lim }\frac{2x^3{\left(\frac{\sin 2x^2}{2x^2}\right)}^2{\cos }^{2}2x}{{\left(\frac{\sin 2x}{2x}\right)}^2}$$$$=\frac{2\times 0\times 1\times 1}{1}=0$$

Commented by ridwan balatif last updated on 22/Jan/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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