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Question Number 122824 by bemath last updated on 19/Nov/20

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

Commented by bobhans last updated on 20/Nov/20

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

Answered by liberty last updated on 19/Nov/20

$$\underset{x\to 0}{\lim }\frac{\sin (1-\cos x)}{2x^2}=\underset{x\to 0}{\lim }\frac{\sin (1-\cos x)}{1-\cos x}.\underset{x\to 0}{\lim }\frac{1-\cos x}{2x^2}$$$$thefirstlimit\underset{x\to 0}{\lim }\frac{\sin (1-\cos x)}{1-\cos x}=\underset{X\to 0}{\lim }\frac{\sin X}{X}=1$$$$thesecondlimit\underset{x\to 0}{\lim }\frac{1-\cos x}{2x^2}=\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{x}{2}\right)}{2x^2}=\frac{1}{4}$$$$thus\underset{x\to 0}{\lim }\frac{\sin (1-\cos x)}{2x^2}=\frac{1}{4}.▴$$

Answered by mathmax by abdo last updated on 20/Nov/20

$$1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow \sin (1-\mathrm{cosx})\sim \sin \left(\frac{{\mathrm{x}}^2}{2}\right)\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow$$$$\frac{\sin (1-\mathrm{cosx})}{{2\mathrm{x}}^2}\sim \frac{{\mathrm{x}}^2}{2\left({2\mathrm{x}}^2\right)}=\frac{1}{4}\Rightarrow {\lim }_{\mathrm{x}\to 0}\frac{\sin (1-\mathrm{cosx})}{{2\mathrm{x}}^2}=\frac{1}{4}$$

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