Question Number 178178 by cortano1 last updated on 13/Oct/22
Answered by Frix last updated on 14/Oct/22
![lim_(x→0) ((x−sin^(−1) x)/(x^2 sin^(−1) x)) =_([t=sin^(−1) x]) =−lim_(t→0) ((t−sin t)/(tsin^2 t)) =_([3 times l′Ho^ pital]) =−lim_(t→0) (((d^3 [t−sin t])/dt^3 )/((d^3 [tsin^2 t])/dt^3 )) = =−lim_(t→0) ((cos t)/(12cos^2 t −8tcos t sin t −6)) = =−(1/6)](https://www.tinkutara.com/question/Q178210.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}^{−\mathrm{1}} \:{x}}{{x}^{\mathrm{2}} \mathrm{sin}^{−\mathrm{1}} \:{x}}\:=_{\left[{t}=\mathrm{sin}^{−\mathrm{1}} \:{x}\right]} \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}−\mathrm{sin}\:{t}}{{t}\mathrm{sin}^{\mathrm{2}} \:{t}}\:=_{\left[\mathrm{3}\:\mathrm{times}\:\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right]} \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}^{\mathrm{3}} \left[{t}−\mathrm{sin}\:{t}\right]}{{dt}^{\mathrm{3}} }}{\frac{{d}^{\mathrm{3}} \left[{t}\mathrm{sin}^{\mathrm{2}} \:{t}\right]}{{dt}^{\mathrm{3}} }}\:= \\ $$$$=−\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{t}}{\mathrm{12cos}^{\mathrm{2}} \:{t}\:−\mathrm{8}{t}\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}\:−\mathrm{6}}\:= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{6}} \\ $$