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Question Number 171341 by mathlove last updated on 13/Jun/22

lim_(x→0) ((x^2 +sin3x)/(2x−sinx))=?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} +{sin}\mathrm{3}{x}}{\mathrm{2}{x}−{sinx}}=? \\ $$

Commented by infinityaction last updated on 13/Jun/22

lim_(x→0) ((x(x+((3sin 3x)/(3x))))/(x(2−((sin x)/x))))      ((0+3)/(2−1)) = 3

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\cancel{{x}}\left({x}+\frac{\mathrm{3sin}\:\mathrm{3}{x}}{\mathrm{3}{x}}\right)}{\cancel{{x}}\left(\mathrm{2}−\frac{\mathrm{sin}\:{x}}{{x}}\right)} \\ $$$$\:\:\:\:\frac{\mathrm{0}+\mathrm{3}}{\mathrm{2}−\mathrm{1}}\:=\:\mathrm{3} \\ $$

Answered by Mathspace last updated on 13/Jun/22

sin(3x)∼3x and sinx∼x ⇒  ((x^2 +3x)/(2x−sinx))∼((x^2 +3x)/(2x−x))=((x(x+3))/x)  =x+3 ⇒  lim_(x→0) ((x^2 +sin(3x))/(2x−sinx))=3

$${sin}\left(\mathrm{3}{x}\right)\sim\mathrm{3}{x}\:{and}\:{sinx}\sim{x}\:\Rightarrow \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{2}{x}−{sinx}}\sim\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}}{\mathrm{2}{x}−{x}}=\frac{{x}\left({x}+\mathrm{3}\right)}{{x}} \\ $$$$={x}+\mathrm{3}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}^{\mathrm{2}} +{sin}\left(\mathrm{3}{x}\right)}{\mathrm{2}{x}−{sinx}}=\mathrm{3} \\ $$

Commented by mathlove last updated on 13/Jun/22

thanks

$${thanks} \\ $$

Answered by nurtani last updated on 13/Jun/22

lim_(x→0)  ((x^2 + sin 3x)/(2x−sin x)) = lim_(x→0)  ((2x+3 cos 3x)/(2−cos x))                                       = ((2(0)+3 cos 3(0))/(2−cos 0))                                       = ((0+3)/(2−1))                                       = 3

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} +\:{sin}\:\mathrm{3}{x}}{\mathrm{2}{x}−{sin}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}+\mathrm{3}\:{cos}\:\mathrm{3}{x}}{\mathrm{2}−{cos}\:{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}\left(\mathrm{0}\right)+\mathrm{3}\:{cos}\:\mathrm{3}\left(\mathrm{0}\right)}{\mathrm{2}−{cos}\:\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{0}+\mathrm{3}}{\mathrm{2}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{3} \\ $$

Commented by mathlove last updated on 13/Jun/22

with out Hspital ruls

$${with}\:{out}\:{Hspital}\:{ruls} \\ $$

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