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Question Number 105584 by bemath last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }[{\csc }^2\left(2x\right)-\frac{1}{4x^2}]?$$

Answered by bobhans last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }[\frac{1}{\sin {}^2\left(2x\right)}-\frac{1}{4x^2}]=\underset{x\to 0}{\lim }\frac{(2x+\sin 2x)(2x-\sin 2x)}{4x^2\sin {}^2\left(2x\right)}$$$$=\underset{x\to 0}{\lim }\frac{(2x+(2x-\frac{{\left(2x\right)}^3}{6}))(2x-(2x-\frac{{\left(2x\right)}^3}{6}))}{\left(4x^2\right)\left(4x^2\right)}$$$$=\underset{x\to 0}{\lim }\frac{(4x-\frac{4x^3}{3})\left(\frac{4x^3}{3}\right)}{16x^4}=\underset{x\to 0}{\lim }\frac{4x(1-\frac{1}{3}{x}^2)}{x}$$$$\times \underset{x\to 0}{\lim }\frac{4x^3}{3.16x^3}=4\times \frac{1}{12}=\frac{1}{3}.$$

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