lim-x-0-x-sin-x-x-2-x-

Question Number 106175 by john santu last updated on 03/Aug/20

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}\:=\:? \\ $$

Answered by bemath last updated on 03/Aug/20

lim_(x→0) (((√x)−(√(x−(x^3 /6))))/(x^2 (√x))) = lim_(x→0) (((√x) (1−(√(1−(x^2 /6)))))/(x^2 (√x))) = lim_(x→0) ((1−(1−(x^2 /(12))))/x^2 ) = (1/(12)) ■

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{x}}\:\left(\mathrm{1}−\sqrt{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}}\right)}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{12}}\right)}{\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{12}}\:\blacksquare\: \\ $$

Answered by mathmax by abdo last updated on 03/Aug/20

let f(x) =(((√x)−(√(sinx)))/(x^2 (√x))) ⇒f(x) =((1−(√((sinx)/x)))/x^2 ) we have sinx ∼x−(x^3 /6) ⇒((sinx)/x)∼1−(x^2 /6) ⇒(√((sinx)/x))∼(√(1−(x^2 /6))) ∼1−(x^2 /(12)) ⇒f(x)∼((1−(1−(x^2 /(12))))/x^2 ) =(1/(12)) ⇒lim_(x→0) f(x) =(1/(12))

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{sinx}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}−\sqrt{\frac{\mathrm{sinx}}{\mathrm{x}}}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\frac{\mathrm{sinx}}{\mathrm{x}}\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow\sqrt{\frac{\mathrm{sinx}}{\mathrm{x}}}\sim\sqrt{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{12}}\right)}{\mathrm{x}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{12}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Answered by Dwaipayan Shikari last updated on 03/Aug/20

lim_(x→0) ((x−sinx)/( (√x)+(√)sinx)).(1/(x^2 (√x)))=lim_(x→0) ((1−((sinx)/x))/(2(√x))).(1/(x(√x)))=((1−1+(x^2 /6))/(2x^2 ))=(1/(12)) (√(sinx))→(√x)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{sinx}}{\:\sqrt{{x}}+\sqrt{}{sinx}}.\frac{\mathrm{1}}{{x}^{\mathrm{2}} \sqrt{{x}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{{sinx}}{{x}}}{\mathrm{2}\sqrt{{x}}}.\frac{\mathrm{1}}{{x}\sqrt{{x}}}=\frac{\mathrm{1}−\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{6}}}{\mathrm{2}{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\sqrt{{sinx}}\rightarrow\sqrt{{x}} \\ $$

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