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Question Number 141900 by iloveisrael last updated on 24/May/21

$$\mathcal{L}=\underset{x\to 0}{\lim }\frac{e^x\cos x-x-1}{x^3}=?$$

Answered by Dwaipayan Shikari last updated on 24/May/21

$$\underset{x\to 0}{\lim }\frac{e^x\left(cosx\right)-x-1}{x^3}=\frac{e^x(1-\frac{x^2}{2})-x-1}{x^3}$$$$=\frac{e^x-x-1-e^x\frac{x^2}{2}}{x^3}=\frac{\frac{x^2}{2}+\frac{x^3}{3!}+O\left(x^4\right)-(1+x+O\left(x^2\right))\frac{x^2}{2}}{x^3}$$$$O\left(x^n\right)\to 0\Rightarrow \frac{\frac{x^3}{3!}-\frac{x^3}{3}}{x^3}=-\frac{1}{6}$$

Answered by mathmax by abdo last updated on 25/May/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-\mathrm{x}-1}{{\mathrm{x}}^3}\mathrm{we}\mathrm{have}{\mathrm{e}}^{\mathrm{x}}\sim 1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2}\mathrm{and}\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\Rightarrow$$$${\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\sim (1+\mathrm{x}+\frac{{\mathrm{x}}^2}{2})(1-\frac{{\mathrm{x}}^2}{2})=1-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}-\frac{{\mathrm{x}}^3}{2}+\frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{x}}^4}{4}$$$$=1+\mathrm{x}-\frac{{\mathrm{x}}^3}{2}-\frac{{\mathrm{x}}^4}{4}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{1+\mathrm{x}-\frac{{\mathrm{x}}^3}{2}-\frac{{\mathrm{x}}^4}{4}-\mathrm{x}-1}{{\mathrm{x}}^3}=-\frac{1}{2}-\frac{\mathrm{x}}{4}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=-\frac{1}{2}$$

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