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EXERCISE 8.1 · Q18

Q.Show that the following function has continuous extension to the point where f(x)f(x) is not defined. Also find the extension: f(x)=1−cos⁡2xsin⁡xf(x) = \dfrac{1-\cos 2x}{\sin x}, for x≠0x \ne 0.

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f(x)=1−cos⁡2xsin⁡xf(x)=\dfrac{1-\cos2x}{\sin x} for x≠0x\ne0; f(0)f(0) is not defined since sin⁡0=0\sin0=0.

Using 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x: for x≠0x\ne0 (so sin⁡x≠0\sin x\ne0 nearby), f(x)=2sin⁡2xsin⁡x=2sin⁡xf(x)=\dfrac{2\sin^2x}{\sin x}=2\sin x.

lim⁡x→0f(x)=lim⁡x→02sin⁡x=2sin⁡0=0.\lim_{x\to0} f(x)=\lim_{x\to0}2\sin x=2\sin0=0. …

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