math-lim-x-pi-sin-2-x-cos-3x-1-

Question Number 109728 by bemath last updated on 25/Aug/20

△((♭ε)/(math))▽ lim_(x→π) ((sin^2 x)/(cos 3x+1))

$$\:\:\:\bigtriangleup\frac{\flat\epsilon}{{math}}\bigtriangledown \\ $$$$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:\mathrm{3}{x}+\mathrm{1}} \\ $$

Commented by bemath last updated on 25/Aug/20

$${thank}\:{you}\:{all}\:{master} \\ $$

Answered by Her_Majesty last updated on 25/Aug/20

=−lim_(x→π) ((2sinxcosx)/(3sin3x)) =lim_(x→π) ((2−4cos^2 x)/(9cos3x)) =(2/9)

$$=−{lim}_{{x}\rightarrow\pi} \frac{\mathrm{2}{sinxcosx}}{\mathrm{3}{sin}\mathrm{3}{x}} \\ $$$$={lim}_{{x}\rightarrow\pi} \frac{\mathrm{2}−\mathrm{4}{cos}^{\mathrm{2}} {x}}{\mathrm{9}{cos}\mathrm{3}{x}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{9}} \\ $$

Answered by bobhans last updated on 25/Aug/20

Answered by 1549442205PVT last updated on 25/Aug/20

Put π−t=x we have sinx=sint cos3x=cos(3π−3t)=cos(π−3t)=−cos3t.Hence lim _(x→π) ((sin^2 x)/(cos 3x+1))=lim_(t→0) ((sin^2 t)/(−cos3t+1))= = _(L′Hopital) lim_(t→0) ((2sintcost)/(3sin3t))=lim_(t→0) ((2sintcost)/(3(3sint−4sin^3 t))) =lim_(t→0) ((2cost)/(3(3−4sin^2 t)))=((2.1)/(3(3−4.0)))=(2/9)

$$\mathrm{Put}\:\pi−\mathrm{t}=\mathrm{x}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}=\mathrm{sint} \\ $$$$\mathrm{cos3x}=\mathrm{cos}\left(\mathrm{3}\pi−\mathrm{3t}\right)=\mathrm{cos}\left(\pi−\mathrm{3t}\right)=−\mathrm{cos3t}.\mathrm{Hence} \\ $$$$\underset{\mathrm{x}\rightarrow\pi} {\mathrm{lim}\:}\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:\mathrm{3}{x}+\mathrm{1}}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{−\mathrm{cos3t}+\mathrm{1}}= \\ $$$$\underset{\mathrm{L}'\mathrm{Hopital}} {=\:\:\:\:}\:\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sintcost}}{\mathrm{3sin3t}}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2sintcost}}{\mathrm{3}\left(\mathrm{3sint}−\mathrm{4sin}^{\mathrm{3}} \mathrm{t}\right)} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\frac{\mathrm{2}\boldsymbol{\mathrm{cost}}}{\mathrm{3}\left(\mathrm{3}−\mathrm{4}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{t}}\right)}=\frac{\mathrm{2}.\mathrm{1}}{\mathrm{3}\left(\mathrm{3}−\mathrm{4}.\mathrm{0}\right)}=\frac{\mathrm{2}}{\mathrm{9}}\: \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 25/Aug/20

let f(x) =((sin^2 x)/(1+cos(3x))) changement x =t+π give f(x) =((sin^2 t)/(1+cos(3t+π))) =((sin^2 t)/(1−cos(3t))) =g(t) (t→0) cos(3t) ∼1−((9t^2 )/2) ⇒1−cos(3t)∼((9t^2 )/2) and sin^2 t ∼ t^2 ⇒ g(t) ∼(2/9) ⇒lim_(x→π) f(x) =(2/9)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{3x}\right)}\:\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{t}+\pi\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{3t}+\pi\right)}\:=\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{t}}{\mathrm{1}−\mathrm{cos}\left(\mathrm{3t}\right)}\:=\mathrm{g}\left(\mathrm{t}\right)\:\:\left(\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{cos}\left(\mathrm{3t}\right)\:\sim\mathrm{1}−\frac{\mathrm{9t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{1}−\mathrm{cos}\left(\mathrm{3t}\right)\sim\frac{\mathrm{9t}^{\mathrm{2}} }{\mathrm{2}}\:\:\mathrm{and}\:\mathrm{sin}^{\mathrm{2}} \mathrm{t}\:\sim\:\mathrm{t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{t}\right)\:\sim\frac{\mathrm{2}}{\mathrm{9}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\pi} \mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{9}} \\ $$

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