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Question Number 130852 by Algoritm last updated on 29/Jan/21

Answered by Dwaipayan Shikari last updated on 29/Jan/21

$$\underset{x\to 0}{\lim }\frac{1-cos\frac{x}{2}}{x^2}=2\left(\frac{{sin}^2\frac{x}{4}}{x^2}\right)=\frac{2}{16}=\frac{1}{8}$$

Answered by benjo_mathlover last updated on 30/Jan/21

$${\mathrm{L}}^{\prime }\mathrm{H}\stackrel{,,}{\mathrm{o}pital}\Rightarrow \underset{x\to 0}{\lim }\frac{\frac{1}{2}\sin \left(\frac{\mathrm{x}}{2}\right)}{2\mathrm{x}}=\frac{\frac{1}{4}}{2}=\frac{1}{8}$$

Answered by malwan last updated on 30/Jan/21

$$\underset{x\to 0}{lim}\frac{1-[1-\frac{{\left(\frac{x}{2}\right)}^2}{2!}+\frac{{\left(\frac{x}{2}\right)}^4}{4!}-...]}{x^2}$$$$=\frac{\left(\frac{1}{4}\right)}{2\times 1}=\frac{1}{8}$$

Answered by malwan last updated on 30/Jan/21

$$anotherway$$$$\underset{x\to 0}{lim}\frac{1-{cos}^2\frac{x}{2}}{x^2(1+cos\frac{x}{2})}=\underset{x\to 0}{lim}\frac{{sin}^2\frac{x}{2}}{2x^2}$$$$=\frac{\frac{1}{2}\times \frac{1}{2}}{2}=\frac{1}{8}$$

Answered by malwan last updated on 30/Jan/21

$$anothersolution$$$$\underset{x\to 0}{lim}\frac{cox0-cos\frac{x}{2}}{x^2}=\underset{x\to 0}{lim}\frac{-2sin\frac{x}{4}sin\frac{-x}{4}}{x^2}$$$$=\frac{-2\times \frac{1}{4}\times \frac{-1}{4}}{1^2}=\frac{1}{8}$$

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