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Question Number 114259 by bobhans last updated on 18/Sep/20

lim_(x→0) ((x arc sin (x^2 ))/(x cos x−sin x)) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}\:? \\ $$

Answered by bemath last updated on 18/Sep/20

by L′Hopital lim_(x→0) ((arc sin (x^2 )+((2x^2 )/( (√(1−x^4 )))))/(cos x−x sin x −cos x)) = lim_(x→0) ((((arc sin (x^2 ))/x^2 ) + (2/( (√(1−x^4 )))))/(−((sin x)/x))) = now let L_1 = lim_(x→0) ((arc sin (x^2 ))/x^2 ) L_1 = lim_(x→0) (((2x)/( (√(1−x^4 ))))/(2x)) = 1 so L = lim_(x→0) ((((arc sin (x^2 ))/x^2 ) + (2/( (√(1−x^4 )))))/(− ((sin x)/x))) L = ((1+2)/(−1)) = −3

$${by}\:{L}'{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{\mathrm{cos}\:{x}−{x}\:\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{−\frac{\mathrm{sin}\:{x}}{{x}}}\:= \\ $$$${now}\:{let}\:{L}_{\mathrm{1}} \:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$${L}_{\mathrm{1}} =\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{\mathrm{2}{x}}\:=\:\mathrm{1} \\ $$$${so}\:{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{arc}\:\mathrm{sin}\:\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}}{−\:\frac{\mathrm{sin}\:{x}}{{x}}} \\ $$$${L}\:=\:\frac{\mathrm{1}+\mathrm{2}}{−\mathrm{1}}\:=\:−\mathrm{3} \\ $$

Answered by Olaf last updated on 18/Sep/20

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

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

Commented by bemath last updated on 18/Sep/20

santuyy

$${santuyy} \\ $$

Commented by malwaan last updated on 18/Sep/20

verysantuyygood

$${verysantuyygood} \\ $$

Commented by abdullahquwatan last updated on 18/Sep/20

mantep

$${mantep} \\ $$

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

we have xarcsin(x^2 )∼x^3 cosx ∼1−(x^2 /2) ⇒xcosx ∼x−(x^3 /2) and sinx ∼x−(x^3 /6) ⇒ f(x)∼(x^3 /(x−(x^3 /2)−x +(x^3 /6))) =(x^3 /(−(1/3)x^3 )) =−3 ⇒lim_(x→0) f(x) =−3

$$\mathrm{we}\:\mathrm{have}\:\mathrm{xarcsin}\left(\mathrm{x}^{\mathrm{2}} \right)\sim\mathrm{x}^{\mathrm{3}} \\ $$$$\mathrm{cosx}\:\sim\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{xcosx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{sinx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\mathrm{x}\:+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}}\:=\frac{\mathrm{x}^{\mathrm{3}} }{−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} }\:=−\mathrm{3}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=−\mathrm{3} \\ $$

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