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Question Number 73948 by Hardy lanes last updated on 17/Nov/19

$$lim\left(\frac{\sin {}^{2}2x}{x}\right)$$$$x\to 0$$

Commented by mathmax by abdo last updated on 17/Nov/19

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

Answered by MJS last updated on 17/Nov/19

$$\underset{x\to 0}{\lim }\frac{{\sin }^{2}2x}{x}=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\left[{\sin }^{2}2x\right]}{\frac{d}{dx}\left[x\right]}=\underset{x\to 0}{\lim }\frac{4\sin 2x\cos 2x}{1}=0$$

Answered by $@ty@m123 last updated on 17/Nov/19

$$=\underset{x\to 0}{\lim }\left(\frac{\sin 2x}{x}\times \sin 2x\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{\sin 2x}{2x}\times 2\times \sin 2x\right)$$$$=2\times \underset{x\to 0}{\lim }\left(\frac{\sin 2x}{2x}\right)\times \underset{x\to 0}{\lim }\left(\sin 2x\right)$$$$=2\times 1\times 0$$$$=0$$

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