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Question Number 117498 by bobhans last updated on 12/Oct/20
lim_(x→0) ((sin (x−sin x))/( (√(1+x^3 ))−1)) ?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }−\mathrm{1}}\:? \\ $$
Answered by bemath last updated on 12/Oct/20
lim_(x→0) ((sin (x−sin x))/( (√(1+x^3 ))−1)) = lim_(x→0) ((x−(x−(x^3 /6)))/((1+(x^3 /2))−1)) = (2/6)×lim_(x→0) (x^3 /x^3 ) = (1/3).
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }−\mathrm{1}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)}{\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}\right)−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{6}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}.\: \\ $$
Answered by Dwaipayan Shikari last updated on 12/Oct/20
lim_(x→0) ((sin(x−sinx))/(1+x^3 −1)).((√(1+x^3 ))+1) lim_(x→0) 2.((x−sinx)/x^3 )=2((x−x+(x^3 /6))/x^3 )=(1/3)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sin}\left({x}−{sinx}\right)}{\mathrm{1}+{x}^{\mathrm{3}} −\mathrm{1}}.\left(\sqrt{\mathrm{1}+{x}^{\mathrm{3}} }+\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}2}.\frac{{x}−{sinx}}{{x}^{\mathrm{3}} }=\mathrm{2}\frac{{x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}}{{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by mathmax by abdo last updated on 12/Oct/20
we have sinx ∼x−(x^3 /6) ⇒sinx−x∼−(x^3 /6) ⇒x−sinx ∼(x^3 /6) and sin(x−sinx)∼sin((x^3 /6))∼(x^3 /6) (1+x^3 )^(1/2) ∼1+(x^3 /2) ⇒(1+x^3 )^(1/2) −1∼(x^3 /2) ⇒ ((sin(x−sinx))/( (√(1+x^3 ))−1))∼((x^3 /6)/(x^3 /2))=(1/3) ⇒lim_(x→0) ((sin(x−sinx))/( (√(1+x^3 ))−1))=(1/3)
$$\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{sinx}−\mathrm{x}\sim−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow\mathrm{x}−\mathrm{sinx}\:\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\mathrm{and}\:\mathrm{sin}\left(\mathrm{x}−\mathrm{sinx}\right)\sim\mathrm{sin}\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\sim\mathrm{1}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}\:\Rightarrow\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\frac{\mathrm{sin}\left(\mathrm{x}−\mathrm{sinx}\right)}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }−\mathrm{1}}\sim\frac{\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}}{\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{sin}\left(\mathrm{x}−\mathrm{sinx}\right)}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{3}} }−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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