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Question Number 66379 by Sandy Suhendra last updated on 13/Aug/19

Commented by kaivan.ahmadi last updated on 13/Aug/19

=lim_(x→(π/2))  ((3cosx+5cos3x)/(cot5x)) =^(hop)   lim_(x→((.π)/2))   ((−3sinx−15sin3x)/(−5(1+cot^2 5x)))=((−3+15)/(−5))=−((12)/5)

$$={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{3}{cosx}+\mathrm{5}{cos}\mathrm{3}{x}}{{cot}\mathrm{5}{x}}\:\overset{{hop}} {=} \\ $$$${lim}_{{x}\rightarrow\frac{.\pi}{\mathrm{2}}} \:\:\frac{−\mathrm{3}{sinx}−\mathrm{15}{sin}\mathrm{3}{x}}{−\mathrm{5}\left(\mathrm{1}+{cot}^{\mathrm{2}} \mathrm{5}{x}\right)}=\frac{−\mathrm{3}+\mathrm{15}}{−\mathrm{5}}=−\frac{\mathrm{12}}{\mathrm{5}} \\ $$

Commented by mathmax by abdo last updated on 14/Aug/19

let A(x)=(3cosx +5cos(3x))tan(5x) changement t=(π/2)−x give  lim_(x→(π/2))   A(x) =lim_(t→0)  (3sint+5cos(((3π)/2)−3t))tan(((5π)/2)−5t)  =lim_(t→0) (3sint −5sin(3t))×(1/(tan(5t)))  we have   sint ∼t     ,sin(3t)∼3t  and tan(5t)∼5t ⇒  ((3sint−5sin(3t))/(tan(5t))) ∼((3t−15t)/(5t)) =−((12)/5) ⇒lim_(x→(π/2))   A(x)=−((12)/5)

$${let}\:{A}\left({x}\right)=\left(\mathrm{3}{cosx}\:+\mathrm{5}{cos}\left(\mathrm{3}{x}\right)\right){tan}\left(\mathrm{5}{x}\right)\:{changement}\:{t}=\frac{\pi}{\mathrm{2}}−{x}\:{give} \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:{A}\left({x}\right)\:={lim}_{{t}\rightarrow\mathrm{0}} \:\left(\mathrm{3}{sint}+\mathrm{5}{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−\mathrm{3}{t}\right)\right){tan}\left(\frac{\mathrm{5}\pi}{\mathrm{2}}−\mathrm{5}{t}\right) \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}} \left(\mathrm{3}{sint}\:−\mathrm{5}{sin}\left(\mathrm{3}{t}\right)\right)×\frac{\mathrm{1}}{{tan}\left(\mathrm{5}{t}\right)}\:\:{we}\:{have}\: \\ $$$${sint}\:\sim{t}\:\:\:\:\:,{sin}\left(\mathrm{3}{t}\right)\sim\mathrm{3}{t}\:\:{and}\:{tan}\left(\mathrm{5}{t}\right)\sim\mathrm{5}{t}\:\Rightarrow \\ $$$$\frac{\mathrm{3}{sint}−\mathrm{5}{sin}\left(\mathrm{3}{t}\right)}{{tan}\left(\mathrm{5}{t}\right)}\:\sim\frac{\mathrm{3}{t}−\mathrm{15}{t}}{\mathrm{5}{t}}\:=−\frac{\mathrm{12}}{\mathrm{5}}\:\Rightarrow{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:{A}\left({x}\right)=−\frac{\mathrm{12}}{\mathrm{5}} \\ $$

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