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Question Number 35726 by abdo mathsup 649 cc last updated on 22/May/18

$$find{lim}_{x\to 0}\frac{1-cos\left(sinx\right)}{x^2}$$

Commented by abdo mathsup 649 cc last updated on 23/May/18

$$weknowthat\frac{1-cosu}{u^2}\sim \frac{1}{2}(u\in v\left(0\right))so$$$$\frac{1-cos\left(sinx\right)}{x^2}=\frac{1-cos\left(sinx\right)}{{sin}^{2}x}.\frac{{sin}^{2}x}{x^2}\sim \frac{1}{2}$$$$because\frac{{sin}^{2}x}{x^2}\sim 1(x\to 0)so$$$${lim}_{x\to 0}\frac{1-cos\left(sinx\right)}{x^2}=\frac{1}{2}.$$

Answered by $@ty@m last updated on 22/May/18

$$\underset{x\to 0}{\lim }\frac{1-cos\left(sinx\right)}{x^2}\times \frac{1+cos\left(sinx\right)}{1+\cos \left(\sin x\right)}$$$$\underset{x\to 0}{\lim }\frac{1-{cos}^2\left(sinx\right)}{x^2}\times \frac{1}{1+\cos \left(\sin x\right)}$$$$\underset{x\to 0}{\lim }\frac{\sin {}^2\left(sinx\right)}{\sin {}^{2}x}\times \frac{\sin {}^{2}x}{x^2}\times \frac{1}{1+\cos \left(\sin x\right)}$$$${\left\{\underset{x\to 0}{\lim }\frac{\sin \left(sinx\right)}{\sin x}\right\}}^2\times {\left\{\underset{x\to 0}{\lim }\frac{\sin x}{x}\right\}}^2\times \underset{x\to 0}{\lim }\frac{1}{1+\cos \left(\sin x\right)}$$$$1^2\times 1^2\times \frac{1}{2}$$$$\frac{1}{2}$$

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