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Question Number 57404 by turbo msup by abdo last updated on 03/Apr/19

$$find{lim}_{x\to 0}\frac{sin\left(\pi \sqrt{cosx}\right)}{x^2}$$

Commented by Smail last updated on 04/Apr/19

$$\underset{x\to 0}{lim}\frac{sin\left(\pi \sqrt{cosx}\right)}{x^2}=\underset{x\to 0}{lim}\frac{{\left(sin\left(\pi \sqrt{cosx}\right)\right)}^{\prime }}{{\left(x^2\right)}^{\prime }}$$$$=\underset{x\to 0}{lim}\frac{\pi \times \frac{-sinx}{2\sqrt{cosx}}\times cos\left(\pi \sqrt{cosx}\right)}{2x}$$$$=\underset{x\to 0}{lim}-\frac{\pi }{4}\times \frac{sinx}{x}\times \frac{cos\left(\pi \sqrt{cosx}\right)}{\sqrt{cosx}}$$$$=\frac{-\pi }{4}\times 1\times \frac{cos\left(\pi \times \sqrt{1}\right)}{\sqrt{1}}=\frac{\pi }{4}$$

Commented by maxmathsup by imad last updated on 04/Apr/19

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

Commented by Smail last updated on 04/Apr/19

$$\sqrt{1-\frac{x^2}{2}}\nsim -\frac{x^2}{2}near0because$$$$forx=0\sqrt{1-\frac{0}{2}}=1\nsim 0$$

Commented by Smail last updated on 04/Apr/19

Commented by maxmathsup by imad last updated on 04/Apr/19

$$looksirihavewritten\sqrt{1-\frac{x^2}{2}}\sim -\frac{x^2}{4}(x\to 0)because\sqrt{1-u}\sim 1-\frac{u}{2}.$$

Answered by kaivan.ahmadi last updated on 04/Apr/19

$$\sim {lim}_{x\to 0}\frac{\pi \sqrt{cosx}}{x^2}andhop$$$$={lim}_{x\to 0}\frac{\pi \frac{-sinx}{2\sqrt{cosx}}}{2x}\sim {lim}_{x\to 0}\frac{-\pi }{4\sqrt{cosx}}=\frac{-\pi }{4}$$

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