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Question Number 60386 by Kunal12588 last updated on 20/May/19

$$\underset{x\to 0}{lim}\frac{sin\left(\pi {cos}^{2}x\right)}{x^2}$$$$whyitcannotbesolvedthisway$$$$\underset{x\to 0}{lim}\frac{sin\left(\pi {cos}^{2}x\right)}{x^2}$$$$=\underset{x\to 0}{lim}\frac{sin\left(\pi {cos}^{2}x\right)}{\pi {cos}^{2}x}\times \underset{x\to 0}{lim}\frac{\pi {cos}^{2}x}{x^2}$$$$=\pi \times \underset{x\to 0}{lim}\frac{{cos}^{2}x}{x^2}$$$$butitisnotequalto\pi$$

Commented by mr W last updated on 20/May/19

$$\underset{x\to 0}{lim}\frac{sin\left(\pi {cos}^{2}x\right)}{\pi {cos}^{2}x}=\frac{\sin \pi }{\pi \times 1}=\frac{0}{\pi }=0\ne \pi$$$$\underset{x\to 0}{\lim }\frac{\pi {cos}^{2}x}{x^2}=\frac{\pi }{0}\to \mathrm{\infty }$$

Commented by mr W last updated on 20/May/19

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

Commented by prakash jain last updated on 20/May/19

Limit of product of two function = product of limits this is only true if both limit exist and finite.

Commented by Mr X pcx last updated on 20/May/19

$$anotherwaywehsvecosx\sim 1-\frac{x^2}{2}\Rightarrow$$$${cos}^{2}x\sim 1-x^2+\frac{x^4}{4}\sim 1-x^2+o\left(x^4\right)\Rightarrow$$$$\pi {cos}^{2}x\sim \pi -\pi x^2\Rightarrow \frac{sin\left(\pi {cos}^{2}x\right)}{x^2}$$$$\sim \frac{sin\left(\pi x^2\right)}{x^2}\sim \frac{\pi x^2}{x^2}(=\pi )\Rightarrow$$$${lim}_{x-0}\frac{sin\left(\pi {cos}^{2}x\right)}{\stackrel{2}{x}}=\pi .$$

Answered by mr W last updated on 20/May/19

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

Commented by Prithwish sen last updated on 20/May/19

$$\mathrm{Sir}\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{will}\mathrm{be}$$$${\lim }_{\mathrm{x}\to 0}\frac{\sin (\pi -\pi {\cos }^2\mathrm{x})}{{\mathrm{x}}^2}\left(2^{\mathrm{nd}}\mathrm{line}\right)$$$$\mathrm{please}\mathrm{check}.$$

Commented by Kunal12588 last updated on 20/May/19

$$yessiriknowtheanswerthiswaybut$$$$whyitcannotbesolvedthewayihavewritten$$

Commented by mr W last updated on 20/May/19

$${\cos }^{2}x=1-{\sin }^{2}x$$$$\Rightarrow \pi {\cos }^{2}x=\pi -\pi {\sin }^{2}x$$$$\Rightarrow \sin \left(\pi {\cos }^{2}x\right)=\sin (\pi -\pi {\sin }^{2}x)$$

Commented by Prithwish sen last updated on 21/May/19

$$\mathrm{Yes}\mathrm{sir}\mathrm{you}\mathrm{are}\mathrm{right}.\mathrm{Please}\mathrm{forgive}.$$

Commented by mr W last updated on 22/May/19

$$noproblemsir!$$

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