Question Number 171341 by mathlove last updated on 13/Jun/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} +{sin}\mathrm{3}{x}}{\mathrm{2}{x}−{sinx}}=? \\ $$
Commented by infinityaction last updated on 13/Jun/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\cancel{{x}}\left({x}+\frac{\mathrm{3sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}\right)}{\cancel{{x}}\left(\mathrm{2}−\frac{\mathrm{sin}\:{x}}{{x}}\right)} \\ $$$$\:\:\:\:\frac{\mathrm{0}+\mathrm{3}}{\mathrm{2}−\mathrm{1}}\:=\:\mathrm{3} \\ $$
Answered by Mathspace last updated on 13/Jun/22
$${sin}\left(\mathrm{3}{x}\right)\sim\mathrm{3}{x}\:{and}\:{sinx}\sim{x}\:\Rightarrow \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{2}{x}−{sinx}}\sim\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{2}{x}−{x}}=\frac{{x}\left({x}+\mathrm{3}\right)}{{x}} \\ $$$$={x}+\mathrm{3}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}^{\mathrm{2}} +{sin}\left(\mathrm{3}{x}\right)}{\mathrm{2}{x}−{sinx}}=\mathrm{3} \\ $$
Commented by mathlove last updated on 13/Jun/22
$${thanks} \\ $$
Answered by nurtani last updated on 13/Jun/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} +\:{sin}\:\mathrm{3}{x}}{\mathrm{2}{x}−{sin}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}+\mathrm{3}\:{cos}\:\mathrm{3}{x}}{\mathrm{2}−{cos}\:{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}\left(\mathrm{0}\right)+\mathrm{3}\:{cos}\:\mathrm{3}\left(\mathrm{0}\right)}{\mathrm{2}−{cos}\:\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{0}+\mathrm{3}}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3} \\ $$
Commented by mathlove last updated on 13/Jun/22
$${with}\:{out}\:{Hspital}\:{ruls} \\ $$