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Question Number 104899 by bramlex last updated on 24/Jul/20

$$\underset{x\to 0}{\lim }\frac{\cos \left(\sin x\right)-\cos \left(x\right)}{x^4}?$$

Answered by john santu last updated on 24/Jul/20

$$\underset{x\to 0}{\lim }\frac{-2\sin \left(\frac{x+\sin x}{2}\right)\sin \left(\frac{\sin x-x}{2}\right)}{x^4}$$$$\underset{x\to 0}{\lim }\frac{-2\left(\frac{x+\sin x}{2}\right)\left(\frac{\sin x-x}{2}\right)}{x^4}$$$$\underset{x\to 0}{\lim }-\frac{(x+x+\frac{x^3}{3!})(x+\frac{x^3}{3!}-x)}{2x^4}$$$$\underset{x\to 0}{\lim }-\frac{\frac{x^3}{6}(2x+\frac{x^3}{6})}{2x^4}=-\underset{x\to 0}{\lim }\frac{x^4(2+\frac{x^2}{6})}{12x^4}$$$$=-\frac{1}{6}.\left(JS\mathrm{♠}\right)$$

Answered by mathmax by abdo last updated on 24/Jul/20

$$\mathrm{we}\mathrm{have}\mathrm{cosx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}=1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{4!}+...$$$$\mathrm{sinx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}=\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+...\Rightarrow$$$$\cos \left(\mathrm{sinx}\right)=\cos (\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+...)=1-\frac{1}{2}{(\mathrm{x}-\frac{{\mathrm{x}}^3}{3!})}^2+\frac{1}{4!}{(\mathrm{x}-\frac{{\mathrm{x}}^3}{3!})}^4+...$$$$=1-\frac{{\mathrm{x}}^2}{2}{(1-\frac{{\mathrm{x}}^2}{3!})}^2+\frac{{\mathrm{x}}^4}{4!}{(1-\frac{{\mathrm{x}}^2}{3!})}^4+....$$$$=1-\frac{{\mathrm{x}}^2}{2}(1-2\frac{{\mathrm{x}}^2}{3!}+\frac{{\mathrm{x}}^4}{{(3!)}^2})+\frac{{\mathrm{x}}^4}{4!}{(1-\frac{{\mathrm{x}}^2}{3!})}^2{(1-\frac{{\mathrm{x}}^2}{3!})}^2+...$$$$=1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{3!}-\frac{{\mathrm{x}}^6}{2{(3!)}^2}+\frac{{\mathrm{x}}^4}{4!}(1-2\frac{{\mathrm{x}}^2}{3!}+\frac{{\mathrm{x}}^4}{{(3!)}^2})(1-2\frac{{\mathrm{x}}^2}{3!}+\frac{{\mathrm{x}}^4}{{(3!)}^2})$$$$=1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{3!}-\frac{{\mathrm{x}}^6}{2{(3!)}^2}+\frac{{\mathrm{x}}^4}{4!}(1-\frac{{2\mathrm{x}}^2}{3!}+\frac{{\mathrm{x}}^4}{{(3!)}^2})-\frac{{2\mathrm{x}}^2}{3!}+...)$$$$\sim 1-\frac{{\mathrm{x}}^2}{2}+(\frac{1}{3!}+\frac{1}{4!}){\mathrm{x}}^4\Rightarrow \cos \left(\mathrm{sinx}\right)-\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}+(\frac{1}{3!}+\frac{1}{4!}){\mathrm{x}}^4-1+\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^4}{4!}$$$$=\frac{{\mathrm{x}}^4}{3!}\Rightarrow {\lim }_{\mathrm{x}\to 0}\frac{\cos \left(\mathrm{sinx}\right)-\mathrm{cosx}}{{\mathrm{x}}^4}=\frac{1}{3!}=\frac{1}{6}$$

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