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 195772 by dimentri last updated on 10/Aug/23

$$\:\:\:\:\cancel{\underline{\underbrace{ }}}\:\cancel{ } \\ $$

Answered by MM42 last updated on 10/Aug/23

lim_(x→(π/3)) ((cos((3x)/2) −2cos((3x)/2)×sin((3x)/2))/(4cos((3x)/2)sin((3x)/2)cos3x)) =lim_(x→(π/3)) ((1 −2sin((3x)/2))/(4sin((3x)/2)cos3x)) = (1/4) ✓ =

$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}\:−\mathrm{2}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}×{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{\mathrm{4}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\mathrm{3}{x}} \\ $$$$={lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}} \:\:\frac{\mathrm{1}\:−\mathrm{2}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{\mathrm{4}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\mathrm{3}{x}}\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\checkmark \\ $$$$= \\ $$

Answered by cortano12 last updated on 10/Aug/23

let ((3x)/2)= u then u→(π/2) L= lim_(u→(π/2)) ((cos u−sin 2u)/(sin 4u)) L= lim_(u→(π/2)) ((cos u(1−2sin u))/(2sin 2u cos 2u)) L = lim_(u→(π/2)) ((cos u(1−2sin u))/(4sin u cos u cos 2u)) L= lim_(u→(π/2)) ((1−2sin u)/(4sin u cos 2u)) = ((1−2)/(4(−1))) = (1/4)

$$\:\:\mathrm{let}\:\frac{\mathrm{3x}}{\mathrm{2}}=\:\mathrm{u}\:\mathrm{then}\:\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}} \\ $$$$\:\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{u}−\mathrm{sin}\:\mathrm{2u}}{\mathrm{sin}\:\mathrm{4u}} \\ $$$$\:\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{u}\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{u}\right)}{\mathrm{2sin}\:\mathrm{2u}\:\mathrm{cos}\:\mathrm{2u}} \\ $$$$\:\:\mathrm{L}\:=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{u}\left(\mathrm{1}−\mathrm{2sin}\:\mathrm{u}\right)}{\mathrm{4sin}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{2u}} \\ $$$$\:\:\mathrm{L}=\:\underset{\mathrm{u}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2sin}\:\mathrm{u}}{\mathrm{4sin}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{2u}} \\ $$$$\:\:\:=\:\frac{\mathrm{1}−\mathrm{2}}{\mathrm{4}\left(−\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: [email protected]