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Question Number 105614 by bemath last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }\frac{\tan (\cos 2x-1)}{2x^2}?$$

Commented by PRITHWISH SEN 2 last updated on 30/Jul/20

$$\mathrm{let}\cos 2\mathrm{x}-1=\mathrm{t}\Rightarrow \mathrm{x}\to 0\mathrm{then}\mathrm{t}\to 0$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{\tan \mathrm{t}}{\mathrm{t}}\underset{x\to 0}{\lim }\frac{\cos 2\mathrm{x}-1}{{2\mathrm{x}}^2}=\underset{x\to 0}{\lim }-{\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}^2=-1$$

Answered by Dwaipayan Shikari last updated on 30/Jul/20

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

Answered by bobhans last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }\frac{\tan (1-\frac{4}{2}{x}^2-1)}{2x^2}=\underset{x\to 0}{\lim }\frac{\tan (-2x^2)}{2x^2}$$$$=-1★$$

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