Question Number 145129 by liberty last updated on 02/Jul/21
Answered by EDWIN88 last updated on 03/Jul/21
![lim_(x→0^+ ) ((sin^2 ((1/( (√(1−x^2 )))) −1))/(cos^2 ((1/( (√(1−x^2 )))) −1)(2sin^2 (((√(2x))/2)))^n )) = a ⇒1×(1/2)× (lim_(x→0^+ ) ((sin ((1/( (√(1−x^2 )))) −1))/(sin^n ((√(x/2))))))^2 = a ⇒(1/2)×(lim_(x→0^+ ) ((cos ((1/( (√(1−x^2 )))) −1).x(1−x^2 )^(−3/2) )/(n sin^(n−1) ((√(x/2)))cos ((√(x/2))).(1/(2(√(2x)))))))^2 =a ⇒(1/2).(1/n^2 ).lim_(x→0^+ ) (((2x(√(2x)))/(sin^(n−1) ((√(x/2))))))^2 = a ⇒(4/n^2 ). lim_(x→0^+ ) (x^3 /(sin^(2n−2) ((√(x/2))))) = a let (√x) = t & t→0 ⇒(4/n^2 ).lim_(t→0) (t^6 /(sin^(2n−2) ((t/( (√2)))))) = a since limit finite , it follows that 2n−2=6 or n = 4 so we get limit becomes ⇒(1/4) .lim_(t→0) [ (t/(sin ((t/( (√2)))))) ]^6 = a ⇒ (1/4)×((√2) )^6 = a ⇒ a=2 . Hence n+a = 4+2= 6](Q145135.png)
$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{1}\right)}{\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{1}\right)\left(\mathrm{2sin}^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{2}{x}}}{\mathrm{2}}\right)\right)^{{n}} \:}\:=\:{a} \\ $$$$\:\Rightarrow\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}×\:\left(\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{1}\right)}{\mathrm{sin}\:^{{n}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)}\right)^{\mathrm{2}} =\:{a} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}×\left(\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\mathrm{1}\right).{x}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{−\mathrm{3}/\mathrm{2}} }{{n}\:\mathrm{sin}\:^{{n}−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)\mathrm{cos}\:\left(\sqrt{\frac{{x}}{\mathrm{2}}}\right).\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{x}}}}\right)^{\mathrm{2}} \:={a} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{{n}^{\mathrm{2}} }.\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left(\frac{\mathrm{2}{x}\sqrt{\mathrm{2}{x}}}{\mathrm{sin}\:^{{n}−\mathrm{1}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)}\right)^{\mathrm{2}} \:=\:{a} \\ $$$$\Rightarrow\frac{\mathrm{4}}{{n}^{\mathrm{2}} }.\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} }{\mathrm{sin}\:^{\mathrm{2}{n}−\mathrm{2}} \left(\sqrt{\frac{{x}}{\mathrm{2}}}\right)}\:=\:{a} \\ $$$$\mathrm{let}\:\sqrt{{x}}\:=\:{t}\:\&\:{t}\rightarrow\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{4}}{{n}^{\mathrm{2}} }.\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{t}^{\mathrm{6}} }{\mathrm{sin}\:^{\mathrm{2}{n}−\mathrm{2}} \left(\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)}\:=\:{a} \\ $$$${since}\:{limit}\:{finite}\:,\:{it}\:{follows} \\ $$$${that}\:\mathrm{2}{n}−\mathrm{2}=\mathrm{6}\:{or}\:{n}\:=\:\mathrm{4}\: \\ $$$${so}\:{we}\:{get}\:{limit}\:{becomes} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\:.\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{{t}}{\mathrm{sin}\:\left(\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)}\:\right]^{\mathrm{6}} =\:{a} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{4}}×\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{6}} \:=\:{a} \\ $$$$\Rightarrow\:{a}=\mathrm{2}\:.\:{Hence}\:{n}+{a}\:=\:\mathrm{4}+\mathrm{2}=\:\mathrm{6} \\ $$