Menu Close

lim-x-1-1-x-cos-1-x-2-




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

Leave a Reply

Your email address will not be published. Required fields are marked *