Question Number 170211 by mathlove last updated on 18/May/22
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{tan}^{−\mathrm{1}} \left(\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} }\right)=? \\ $$$${pleas}\:{solve}\:{this} \\ $$
Answered by aleks041103 last updated on 18/May/22
![1) lim_(x→0) ((arctan x)/x) =^(L′H) lim_(x→0) (1/(1+x^2 ))=1 2) lim_(x→0) ((x−x(√(1−x^2 )))/( (√(1−x^2 ))+x^2 ))=0 3) ⇒lim_(x→0) (((tan^(−1) (((x−x(√(1−x^2 )))/( (√(1−x^2 ))+x^2 ))))/x^3 ))= =[lim_(x→0) (((x−x(√(1−x^2 )))/( (√(1−x^2 ))+x^2 ))/x^3 )][lim_(y→0) ((arctan(y))/y)]= =lim_(x→0) (((x−x(√(1−x^2 )))/( (√(1−x^2 ))+x^2 ))/x^3 ) where y=((x−x(√(1−x^2 )))/( (√(1−x^2 ))+x^2 )) 4) ⇒L=lim_(x→0) ((x(1−(√(1−x^2 ))))/(x^3 (x^2 +(√(1−x^2 )))))= =(lim_(x→0) (1/(x^2 +(√(1−x^2 )))))(lim_(x→0) ((1−(√(1−x^2 )))/x^2 )) =lim_(x→0) ((1−(√(1−x^2 )))/x^2 )=lim_(x→0) (((1−(√(1−x^2 )))(1+(√(1−x^2 ))))/(x^2 (1+(√(1−x^2 )))))= =lim_(x→0) ((1−(1−x^2 ))/(x^2 (1+(√(1−x^2 )))))=lim_(x→0) (1/(1+(√(1−x^2 ))))=(1/2) ⇒Ans. (1/2)](https://www.tinkutara.com/question/Q170216.png)
$$\left.\mathrm{1}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{arctan}\:{x}}{{x}}\:\overset{{L}'{H}} {=}\:\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left.\mathrm{3}\right) \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{tan}^{−\mathrm{1}} \left(\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} }\right)= \\ $$$$=\left[\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }}{{x}^{\mathrm{3}} }\right]\left[\underset{{y}\rightarrow\mathrm{0}} {{lim}}\frac{{arctan}\left({y}\right)}{{y}}\right]= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }}{{x}^{\mathrm{3}} } \\ $$$${where}\:{y}=\frac{{x}−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{4}\right) \\ $$$$\Rightarrow{L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}= \\ $$$$=\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{Ans}.\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 18/May/22
$${thanks} \\ $$