Question Number 132137 by benjo_mathlover last updated on 11/Feb/21
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\left(\mathrm{sin}^{−\mathrm{1}} \mathrm{x}−\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\right)\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} =? \\ $$
Answered by EDWIN88 last updated on 11/Feb/21
![L=lim_(x→0) ((2/x^3 )[(x+(x^3 /6)+(x^5 /(40)))−(x−(x^3 /3)+(x^5 /5))])^(2/x^2 ) L=lim_(x→0) ((2/x^3 )((x^3 /2)−((7x^5 )/(40))))^(2/x^2 ) L=lim_(x→0) (1−((7x^2 )/(20)))^(2/x^2 ) = e^(lim_(x→0) (−((14x^2 )/(20x^2 )))) =e^(−(7/(10)))](https://www.tinkutara.com/question/Q132145.png)
$$\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\left[\left(\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{40}}\right)−\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{5}}\right)\right]\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{3}} }\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{2}}−\frac{\mathrm{7x}^{\mathrm{5}} }{\mathrm{40}}\right)\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{7x}^{\mathrm{2}} }{\mathrm{20}}\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} =\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(−\frac{\mathrm{14x}^{\mathrm{2}} }{\mathrm{20x}^{\mathrm{2}} }\right)} =\mathrm{e}^{−\frac{\mathrm{7}}{\mathrm{10}}} \\ $$