Question Number 90281 by jagoll last updated on 22/Apr/20
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\left(\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} }\:=\:? \\ $$
Commented by jagoll last updated on 22/Apr/20
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}}{\mathrm{2cos}^{−\mathrm{1}} \left(\mathrm{x}\right).\left(\frac{−\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\right)}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)}\:=\: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:×\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mathmax by abdo last updated on 22/Apr/20
$${let}\:{f}\left({x}\right)\:=\frac{\mathrm{1}−\sqrt{{x}}}{\left({arcosx}\right)^{\mathrm{2}} }\:\:{changemrnt}\:\:{arcosx}\:={t}\:{give} \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{1}−\sqrt{{cost}}}{{t}^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\left({x}\rightarrow\mathrm{1}\:\Rightarrow{t}\rightarrow\mathrm{0}\right)\:{let}\:{g}\left({t}\right)=\frac{\mathrm{1}−\sqrt{{cost}}}{{t}^{\mathrm{2}} } \\ $$$$\Rightarrow{g}\left({t}\right)=\frac{\mathrm{1}−{cost}}{{t}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\left.{cost}\right)}\right.}\:{we}\:{know}\:{lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{1}−{cost}}{{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:{g}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:={lim}_{{x}\rightarrow\mathrm{1}} {f}\left({x}\right) \\ $$