Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 165868 by cortano1 last updated on 09/Feb/22

lim_(x→(π/2)) ((cos x)/(sin x−((sin x+cos x))^(1/3) )) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}−\sqrt[{\mathrm{3}}]{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}}\:=? \\ $$

Answered by qaz last updated on 10/Feb/22

lim_(x→(π/2)) ((cos x)/(sin x−((sin x+cos x))^(1/3) )) =lim_(x→0) ((−sin x)/(cos x−((cos x−sin x))^(1/3) )) =lim_(x→0) ((−x)/((1/3)ln(1−sin x))) =−3lim_(x→0) (x/(sin x)) =−3

$$\underset{\mathrm{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}−\sqrt[{\mathrm{3}}]{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}−\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}} \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{x}}{\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)} \\ $$$$=−\mathrm{3}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}}{\mathrm{sin}\:\mathrm{x}} \\ $$$$=−\mathrm{3} \\ $$

Answered by Rohit143Jo last updated on 10/Feb/22

Ans: lim_(x→(π/2)) ((Cos x)/(Sin x − ((Sin x + Cos x))^(1/3) )) = lim_(x→(π/2)) (((−Sin x))/(Cos x − ((Cos x − Sin x)/(3 (Sin x +Cos x)^(2/3) )))) [L′Hospital Rule] = lim_(x→(π/2)) ((3.(−Sin x).(Sin x + Cos x)^(2/3) )/(3Cos x.(Sin x + Cos x)^(2/3) − Cos x + Sin x)) = ((3.(−Sin (π/2)).(Sin (π/2) + Cos (π/2))^(2/3) )/(3.Cos (π/2).(Sin (π/2) + Cos (π/2))^(2/3) − Cos (π/2) + Sin (π/2))) = (−3).

$${Ans}:\:\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\frac{{Cos}\:{x}}{{Sin}\:{x}\:−\:\sqrt[{\mathrm{3}}]{{Sin}\:{x}\:+\:{Cos}\:{x}}} \\ $$$$\:\:\:\:\:\:\:=\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\frac{\left(−{Sin}\:{x}\right)}{{Cos}\:{x}\:−\:\frac{{Cos}\:{x}\:−\:{Sin}\:{x}}{\mathrm{3}\:\left({Sin}\:{x}\:+{Cos}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }}\:\left[{L}'{Hospital}\:{Rule}\right] \\ $$$$\:\:\:\:\:\:\:=\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{3}.\left(−{Sin}\:{x}\right).\left({Sin}\:{x}\:+\:{Cos}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}{Cos}\:{x}.\left({Sin}\:{x}\:+\:{Cos}\:{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\:{Cos}\:{x}\:+\:{Sin}\:{x}}\: \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{\mathrm{3}.\left(−{Sin}\:\frac{\pi}{\mathrm{2}}\right).\left({Sin}\:\frac{\pi}{\mathrm{2}}\:+\:{Cos}\:\frac{\pi}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}.{Cos}\:\frac{\pi}{\mathrm{2}}.\left({Sin}\:\frac{\pi}{\mathrm{2}}\:+\:{Cos}\:\frac{\pi}{\mathrm{2}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:−\:{Cos}\:\frac{\pi}{\mathrm{2}}\:+\:{Sin}\:\frac{\pi}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:=\:\left(−\mathrm{3}\right). \\ $$

Answered by greogoury55 last updated on 15/Feb/22

Terms of Service

Privacy Policy

Contact: info@tinkutara.com