Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 147349 by bramlexs22 last updated on 20/Jul/21

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}(1-\cos \mathrm{x})}{{\mathrm{x}}^2+\mathrm{x}-{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}}=?$$

Commented by bramlexs22 last updated on 20/Jul/21

Commented by EDWIN88 last updated on 20/Jul/21

$$\mathrm{via}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 0}{\lim }\frac{2\mathrm{x}\sin {}^2\left(\frac{\mathrm{x}}{2}\right)}{{\mathrm{x}}^2+\mathrm{x}-{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}}=$$$$\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{\mathrm{x}}{2}\right)+\mathrm{x}\sin \mathrm{x}}{2\mathrm{x}+1-({\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x})}=$$$$\underset{x\to 0}{\lim }\frac{2\left(\frac{1-\cos \mathrm{x}}{2}\right)+\mathrm{x}\sin \mathrm{x}}{2\mathrm{x}+1-({\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x})}=$$$$\underset{x\to 0}{\lim }\frac{1-\cos \mathrm{x}+\mathrm{x}\sin \mathrm{x}}{2\mathrm{x}+1-({\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+{\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x})}=$$$$\underset{x\to 0}{\lim }\frac{2\sin \mathrm{x}+\mathrm{x}\cos \mathrm{x}}{2-({\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x}+{2\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}-{\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x})}=$$$$\underset{x\to 0}{\lim }\frac{3\cos \mathrm{x}-\mathrm{x}\sin \mathrm{x}}{-({2\mathrm{e}}^{\mathrm{x}}\cos \mathrm{x}-{2\mathrm{e}}^{\mathrm{x}}\sin \mathrm{x})}=$$$$-\frac{3}{2}.✓$$

Answered by ArielVyny last updated on 20/Jul/21

$$cosx\sim 1-\frac{x^2}{2}$$$$e^{x}sinx\sim (1-x)x$$$$li\underset{x\to 0}{m}\frac{x(1-cosx)}{x^2+x-e^{x}sinx}=li\underset{x\to 0}{m}\frac{x(1-1+\frac{x^2}{2})}{x^2+x-(x-x^2)}$$$$=li\underset{x\to 0}{m}\frac{\frac{x^3}{2}}{2x^2}=\frac{x^3}{2}\times \frac{1}{2x^2}=\frac{x}{4}=0$$$$$$

Commented by bramlexs22 last updated on 20/Jul/21

$$\mathrm{false}$$

Answered by mathmax by abdo last updated on 20/Jul/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}(1-\mathrm{cosx})}{{\mathrm{x}}^2+\mathrm{x}-{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}}$$$$\mathrm{we}\mathrm{have}1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2},{\mathrm{e}}^{\mathrm{x}}\sim 1+\mathrm{x},\mathrm{sinx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{6}\Rightarrow$$$${\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\sim (1+\mathrm{x})(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})=\mathrm{x}-\frac{{\mathrm{x}}^3}{6}+{\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{6}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{\frac{{\mathrm{x}}^3}{2}}{{\mathrm{x}}^2+\mathrm{x}-\mathrm{x}+\frac{{\mathrm{x}}^3}{6}-{\mathrm{x}}^2+\frac{{\mathrm{x}}^4}{6}}=\frac{{6\mathrm{x}}^3}{2({\mathrm{x}}^3+{\mathrm{x}}^4)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{3}{1+\mathrm{x}}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=3$$

Commented by liberty last updated on 20/Jul/21

$$ithinktheanswer-\frac{3}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com