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Question Number 100134 by bemath last updated on 25/Jun/20
lim_(x→(π/2)) ((4sin x−(√(6(√(sin x))+10)))/((π/2)−x)) ?
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{4sin}\:\mathrm{x}−\sqrt{\mathrm{6}\sqrt{\mathrm{sin}\:\mathrm{x}}+\mathrm{10}}}{\frac{\pi}{\mathrm{2}}−\mathrm{x}}\:? \\ $$
Commented by bobhans last updated on 25/Jun/20
set (π/2)−x = t , x =(π/2)−t lim_(t→0) ((4cos t−(√(6(√(cos t))+10)))/t) = lim_(t→0) ((16cos^2 t−(6(√(cos t))+10))/(8t)) = lim_(t→0) ((−16sin 2t−((6(−sin t))/(2(√(cos t)))))/8) = 0
$$\mathrm{set}\:\frac{\pi}{\mathrm{2}}−\mathrm{x}\:=\:\mathrm{t}\:,\:\mathrm{x}\:=\frac{\pi}{\mathrm{2}}−\mathrm{t}\: \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4cos}\:\mathrm{t}−\sqrt{\mathrm{6}\sqrt{\mathrm{cos}\:\mathrm{t}}+\mathrm{10}}}{\mathrm{t}}\:= \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{16cos}\:^{\mathrm{2}} \mathrm{t}−\left(\mathrm{6}\sqrt{\mathrm{cos}\:\mathrm{t}}+\mathrm{10}\right)}{\mathrm{8t}}\:= \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{16sin}\:\mathrm{2t}−\frac{\mathrm{6}\left(−\mathrm{sin}\:\mathrm{t}\right)}{\mathrm{2}\sqrt{\mathrm{cos}\:\mathrm{t}}}}{\mathrm{8}}\:=\:\mathrm{0} \\ $$$$ \\ $$
Commented by Dwaipayan Shikari last updated on 25/Jun/20
lim_(x→(π/2)) ((−4cosx−(((1/2).(3/( (√(sinx))))cosx)/( (√(6(√(sinx))+10)))))/(−1))=0
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{−\mathrm{4cosx}−\frac{\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{3}}{\:\sqrt{\mathrm{sinx}}}\mathrm{cosx}}{\:\sqrt{\mathrm{6}\sqrt{\mathrm{sinx}}+\mathrm{10}}}}{−\mathrm{1}}=\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 25/Jun/20
changement (π/2)−x =t give f(x)=((4sinx−(√(6(√(sinx))+10)))/((π/2)−x)) =((4cost −(√(6(√(cost))+10)))/t)=g(t) we have cost ∼1−(t^2 /2) ⇒(√(cost))∼1−(t^2 /4) ⇒ 6(√(cost))∼6−(3/2)t^2 ⇒(√(6(√(cost))+10))∼(√(16−(3/2)t^2 ))=4(√(1−((3t^2 )/(32))))∼4(1−((3t^2 )/(64))) ⇒ g(t) ∼((4−2t^2 −4+((3t^2 )/(16)))/t) =(−2+(3/(16)))t →0 ⇒lim_(x→(π/2)) f(x) =0
$$\mathrm{changement}\:\frac{\pi}{\mathrm{2}}−\mathrm{x}\:=\mathrm{t}\:\mathrm{give}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4sinx}−\sqrt{\mathrm{6}\sqrt{\mathrm{sinx}}+\mathrm{10}}}{\frac{\pi}{\mathrm{2}}−\mathrm{x}} \\ $$$$=\frac{\mathrm{4cost}\:−\sqrt{\mathrm{6}\sqrt{\mathrm{cost}}+\mathrm{10}}}{\mathrm{t}}=\mathrm{g}\left(\mathrm{t}\right)\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{cost}\:\sim\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\sqrt{\mathrm{cost}}\sim\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{6}\sqrt{\mathrm{cost}}\sim\mathrm{6}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \:\:\Rightarrow\sqrt{\mathrm{6}\sqrt{\mathrm{cost}}+\mathrm{10}}\sim\sqrt{\mathrm{16}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{1}−\frac{\mathrm{3t}^{\mathrm{2}} }{\mathrm{32}}}\sim\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{3t}^{\mathrm{2}} }{\mathrm{64}}\right)\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{t}\right)\:\sim\frac{\mathrm{4}−\mathrm{2t}^{\mathrm{2}} −\mathrm{4}+\frac{\mathrm{3t}^{\mathrm{2}} }{\mathrm{16}}}{\mathrm{t}}\:=\left(−\mathrm{2}+\frac{\mathrm{3}}{\mathrm{16}}\right)\mathrm{t}\:\rightarrow\mathrm{0}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{0} \\ $$

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