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Question Number 114079 by bemath last updated on 17/Sep/20

Answered by john santu last updated on 17/Sep/20

$$setting\frac{1}{x}=m\wedge m\to 0$$$$\underset{m\to 0}{\lim }\frac{m^2\sin 4m}{(1-\cos 2m)\sin m}=$$$$\underset{m\to 0}{\lim }\frac{m^2\sin 4m}{2\sin {}^{2}m.\sin m}=\frac{4}{2}=2$$

Answered by Olaf last updated on 17/Sep/20

$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{4}{x}}{(1-(1-\frac{1}{2}{\left(\frac{2}{x}\right)}^2))x^2\frac{1}{x}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{4}{x}}{\left(\frac{1}{2}{\left(\frac{2}{x}\right)}^2\right)x^2\frac{1}{x}}$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{4}{x}}{\left(\frac{2}{x^2}\right)x}=\frac{4}{2}=2$$$$$$

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