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Question Number 118681 by bemath last updated on 19/Oct/20

 lim_(x→0)  (((√x) − (√(sin x)))/(x^2  (√x))) =?

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

Answered by benjo_mathlover last updated on 19/Oct/20

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

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:×\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:+\sqrt{\mathrm{sin}\:{x}}}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{3}} }\:×\:\frac{\sqrt{{x}}}{\:\sqrt{{x}}\:+\sqrt{\mathrm{sin}\:{x}}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by Lordose last updated on 19/Oct/20

Very correct

$$\mathrm{Very}\:\mathrm{correct} \\ $$

Commented by 1549442205PVT last updated on 19/Oct/20

 I=lim _(x→0) (((√x) − (√(sin x)))/(x^2  (√x))) .This is the form (0/0)  =lim_(x→0) ((x−sinx)/x^3 )×((√x)/( (√x)+(√(sinx))))  =lim_(x→0) ((x−sinx)/x^3 )×(1/(1+(√((sinx)/x))))=A×(1/2)  A=^(0/0) lim_(x→0) ((1−cosx)/(3x^2 ))=    ^(0/0)    _(L′Hopital) lim_(x→0) ((sinx)/(6x))  =1/6⇒I=(1/6)×(1/2)=(1/(12))

$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\:\mathrm{I}=\mathrm{lim}\:}\frac{\sqrt{{x}}\:−\:\sqrt{\mathrm{sin}\:{x}}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:.\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{form}\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {=\mathrm{lim}}\frac{\mathrm{x}−\mathrm{sinx}}{\mathrm{x}^{\mathrm{3}} }×\frac{\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{x}}+\sqrt{\mathrm{sinx}}} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {=\mathrm{lim}}\frac{\mathrm{x}−\mathrm{sinx}}{\mathrm{x}^{\mathrm{3}} }×\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\frac{\mathrm{sinx}}{\mathrm{x}}}}=\mathrm{A}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{A}=^{\frac{\mathrm{0}}{\mathrm{0}}} \underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{3x}^{\mathrm{2}} }=\:\:\:\underset{\mathrm{L}'\mathrm{Hopital}} {\:\:^{\frac{\mathrm{0}}{\mathrm{0}}} \:\:\:}\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\mathrm{sinx}}{\mathrm{6x}} \\ $$$$=\mathrm{1}/\mathrm{6}\Rightarrow\mathrm{I}=\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Answered by mathmax by abdo last updated on 19/Oct/20

f(x)=(((√x)−(√(sinx)))/(x^2 (√x)))  changement (√x)=t give  f(x)=f(t^2 ) =((t−(√(sin(t^2 ))))/t^5 ) (x→0 ⇒t→0)   sin(t^2 )∼t^2 −(t^6 /6) ⇒(√(sin(t^2 )))∼t(√(1−(t^4 /6)))∼t(1−(t^4 /(12))) ⇒  f(t^2 )∼((t−t+(t^5 /(12)))/t^5 ) ⇒f(t^2 )∼(1/(12)) ⇒lim_(x→0)   f(x)=(1/(12))

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{sinx}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}}\:\:\mathrm{changement}\:\sqrt{\mathrm{x}}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{t}^{\mathrm{2}} \right)\:=\frac{\mathrm{t}−\sqrt{\mathrm{sin}\left(\mathrm{t}^{\mathrm{2}} \right)}}{\mathrm{t}^{\mathrm{5}} }\:\left(\mathrm{x}\rightarrow\mathrm{0}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\right)\: \\ $$$$\mathrm{sin}\left(\mathrm{t}^{\mathrm{2}} \right)\sim\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{t}^{\mathrm{6}} }{\mathrm{6}}\:\Rightarrow\sqrt{\mathrm{sin}\left(\mathrm{t}^{\mathrm{2}} \right)}\sim\mathrm{t}\sqrt{\mathrm{1}−\frac{\mathrm{t}^{\mathrm{4}} }{\mathrm{6}}}\sim\mathrm{t}\left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{4}} }{\mathrm{12}}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{t}^{\mathrm{2}} \right)\sim\frac{\mathrm{t}−\mathrm{t}+\frac{\mathrm{t}^{\mathrm{5}} }{\mathrm{12}}}{\mathrm{t}^{\mathrm{5}} }\:\Rightarrow\mathrm{f}\left(\mathrm{t}^{\mathrm{2}} \right)\sim\frac{\mathrm{1}}{\mathrm{12}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{12}} \\ $$

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