Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com