Question Number 120511 by bramlexs22 last updated on 01/Nov/20
![lim_(x→0) ((√([ sin^(−1) (2x)]^3 ))/(x sin ((√x)))) ?](https://www.tinkutara.com/question/Q120511.png)
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\left[\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\right]^{\mathrm{3}} }}{{x}\:\mathrm{sin}\:\left(\sqrt{{x}}\right)}\:?\: \\ $$
Answered by Olaf last updated on 01/Nov/20
$$\frac{\sqrt{\mathrm{arcsin}^{\mathrm{3}} \left(\mathrm{2}{x}\right)}}{{x}\mathrm{sin}\left(\sqrt{{x}}\right)}\:\underset{\mathrm{0}^{+} } {\sim}\:\frac{\sqrt{\left(\mathrm{2}{x}\right)^{\mathrm{3}} }}{{x}\sqrt{{x}}}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Answered by john santu last updated on 01/Nov/20
![lim_(x→0) ((√([ sin^(−1) (2x) ]^3 ))/(x sin ((√x)))) = lim_(x→0) (((3/2) (√(sin^(−1) (2x))) .(2/( (√(1−4x^2 )))))/(sin (√x) + ((x cos ((√x)))/(2(√x))))) = lim_(x→0) (((3/( (√(1−4x^2 )))).(√(sin^(−1) (2x))))/(sin (√x) +(1/2)(√x) cos ((√x)))) = 6 lim_(x→0) ((√(sin^(−1) (2x)))/(2sin (√x) +(√x))) = 6 lim_(x→0) ((√(2x))/( 3(√x))) = ((6(√2))/3) = 2(√2) note that { ((lim_(x→0) ((sin x)/x) = 1)),((lim_(x→0) ((sin^(−1) (x))/x) = 1)) :}](https://www.tinkutara.com/question/Q120513.png)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\left[\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\:\right]^{\mathrm{3}} }}{{x}\:\mathrm{sin}\:\left(\sqrt{{x}}\right)}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{3}}{\mathrm{2}}\:\sqrt{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}\:.\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}}{\mathrm{sin}\:\sqrt{{x}}\:+\:\frac{{x}\:\mathrm{cos}\:\left(\sqrt{{x}}\right)}{\mathrm{2}\sqrt{{x}}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{3}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}.\sqrt{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}}{\mathrm{sin}\:\sqrt{{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{x}}\:\mathrm{cos}\:\left(\sqrt{{x}}\right)} \\ $$$$=\:\mathrm{6}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}}{\mathrm{2sin}\:\sqrt{{x}}\:\:+\sqrt{{x}}}\:=\:\mathrm{6}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}{x}}}{\:\mathrm{3}\sqrt{{x}}}\:=\:\frac{\mathrm{6}\sqrt{\mathrm{2}}}{\mathrm{3}}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${note}\:{that}\:\begin{cases}{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\:\mathrm{1}}\\{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left({x}\right)}{{x}}\:=\:\mathrm{1}}\end{cases} \\ $$
Answered by bramlexs22 last updated on 01/Nov/20
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)\sqrt{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}}{{x}\:\mathrm{sin}\:\sqrt{{x}}}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}}}}{\frac{\mathrm{sin}\:\sqrt{{x}}}{\:\sqrt{{x}}}}\:= \\ $$$$\:\mathrm{2}\:×\:\frac{\sqrt{\mathrm{2}}}{\mathrm{1}}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$ \\ $$