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 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

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

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)

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com