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Question Number 135437 by 0731619177 last updated on 13/Mar/21

Answered by EDWIN88 last updated on 13/Mar/21

$${\mathrm{L}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}$$$$\underset{x\to 0}{\lim }\frac{\frac{4\mathrm{x}-2\sin 2\mathrm{x}}{{2\mathrm{x}}^2+\cos 2\mathrm{x}}}{{4\mathrm{x}}^3}=\underset{x\to 0}{\lim }\frac{1}{{2\mathrm{x}}^2+\cos 2\mathrm{x}}.\underset{x\to 0}{\lim }\frac{4\mathrm{x}-2\sin 2\mathrm{x}}{{4\mathrm{x}}^3}$$$$=\frac{1}{1}.\underset{x\to 0}{\lim }\frac{4-4\cos 2\mathrm{x}}{{12\mathrm{x}}^2}=\frac{1}{3}.\underset{x\to 0}{\lim }\frac{1-(1-2\sin {}^2\mathrm{x})}{{\mathrm{x}}^2}$$$$=\frac{2}{3}.{\left[\underset{x\to 0}{\lim }\frac{\sin \mathrm{x}}{\mathrm{x}}\right]}^2=\frac{2}{3}$$

Answered by mathmax by abdo last updated on 13/Mar/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\ln ({2\mathrm{x}}^2+\cos \left(2\mathrm{x}\right))}{{\mathrm{x}}^4}\mathrm{we}\mathrm{have}\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{4!}\Rightarrow$$$$\cos \left(2\mathrm{x}\right)\sim 1-{2\mathrm{x}}^2+\frac{{16\mathrm{x}}^4}{4.3.2}\Rightarrow \cos \left(2\mathrm{x}\right)\sim 1-{2\mathrm{x}}^2+\frac{{2\mathrm{x}}^4}{3}\Rightarrow$$$${2\mathrm{x}}^2+\cos \left(2\mathrm{x}\right)\sim 1+\frac{{2\mathrm{x}}^4}{3}\Rightarrow \ln ({2\mathrm{x}}^2+\cos \left(2\mathrm{x}\right))\sim \ln (1+\frac{{2\mathrm{x}}^4}{3})\sim \frac{{2\mathrm{x}}^4}{3}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{2}{3}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{2}{3}$$$$$$

Commented by 0731619177 last updated on 13/Mar/21

$$thanks$$

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