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Question Number 51005 by Tawa1 last updated on 23/Dec/18

Commented by maxmathsup by imad last updated on 23/Dec/18

let A(θ) =((sin(θ−(π/6)))/((√3) −2cosθ)) changement θ −(π/6) =x give A(θ) =((sinx)/((√3)−2cos(x+(π/6)))) =((sinx)/((√3)−2(cosx cos((π/6))−sinx sin((π/6))))) = ((sinx)/((√3)−2(((√3)/2)cos−(1/2)sinx))) =((sinx)/((√3)−(√3)cosx +sinx)) but lim_(θ→(π/6)) =lim_(x→0) ((sinx)/((√3)−(√3)cosx+sinx)) sinx ∼ x and cosx ∼1−(x^2 /2) ⇒ ((sinx)/((√3)−(√3)cosx −sinx)) ∼ (x/((√3)−(√3)(1−(x^2 /2))−x)) = (x/(((√3)/2)x^2 −x)) =(1/(((√3)/2)x−1)) →−1 (x→0) ⇒lim_(θ→(π/6)) A(θ) =−1 . error of typo ((sinx)/((√3)−(√3)cosx +sinx)) ∼ (x/((√3)−(√3)(1−(x^2 /2))+x)) =(x/(((√3)/2)x^2 +x)) = = (1/(((√3)/2)x +1)) →1 (x→0) ⇒★lim_(θ→(π/6)) A(θ) = 1 ★

$${let}\:{A}\left(\theta\right)\:=\frac{{sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)}{\sqrt{\mathrm{3}}\:−\mathrm{2}{cos}\theta}\:\:{changement}\:\:\theta\:−\frac{\pi}{\mathrm{6}}\:={x}\:{give} \\ $$$${A}\left(\theta\right)\:=\frac{{sinx}}{\sqrt{\mathrm{3}}−\mathrm{2}{cos}\left({x}+\frac{\pi}{\mathrm{6}}\right)}\:=\frac{{sinx}}{\sqrt{\mathrm{3}}−\mathrm{2}\left({cosx}\:{cos}\left(\frac{\pi}{\mathrm{6}}\right)−{sinx}\:{sin}\left(\frac{\pi}{\mathrm{6}}\right)\right)} \\ $$$$=\:\frac{{sinx}}{\sqrt{\mathrm{3}}−\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cos}−\frac{\mathrm{1}}{\mathrm{2}}{sinx}\right)}\:=\frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:+{sinx}}\:{but}\:{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:={lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}+{sinx}} \\ $$$${sinx}\:\sim\:{x}\:\:\:{and}\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:\frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:−{sinx}}\:\sim\:\frac{{x}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)−{x}} \\ $$$$=\:\frac{{x}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}^{\mathrm{2}} −{x}}\:=\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}−\mathrm{1}}\:\rightarrow−\mathrm{1}\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:\:{A}\left(\theta\right)\:=−\mathrm{1}\:\:\:\:\:\:. \\ $$$${error}\:{of}\:{typo}\:\:\frac{{sinx}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{cosx}\:+{sinx}}\:\sim\:\frac{{x}}{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)+{x}}\:=\frac{{x}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}^{\mathrm{2}} \:+{x}}\:= \\ $$$$=\:\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\:+\mathrm{1}}\:\rightarrow\mathrm{1}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow\bigstar{lim}_{\theta\rightarrow\frac{\pi}{\mathrm{6}}} \:\:\:{A}\left(\theta\right)\:=\:\mathrm{1}\:\bigstar \\ $$

Commented by maxmathsup by imad last updated on 23/Dec/18

let A(x) =((1−cosx(√(cos(2x))))/x^2 ) we have x ∈V(0) cosx ∼1−(x^2 /2) and cos(2x)∼ 1−(((2x)^2 )/2) =1−2x^2 ⇒(√(cos(2x)))∼(√(1−2x^2 )) ∼1+(1/2)(−2x^2 ) =1−x^2 ⇒A(x)∼ ((1−(1−(x^2 /2))(1−x^2 ))/x^2 ) =((1−(1−x^2 −(x^2 /2) +(x^4 /2)))/x^2 ) =((x^2 +(x^2 /2)−(x^4 /2))/x^2 ) =(3/2) −(x^2 /2) →(3/2) (x→0) ⇒ lim_(x→0) A(x) =(3/2) .

$${let}\:{A}\left({x}\right)\:=\frac{\mathrm{1}−{cosx}\sqrt{{cos}\left(\mathrm{2}{x}\right)}}{{x}^{\mathrm{2}} }\:\:\:{we}\:{have}\:{x}\:\in{V}\left(\mathrm{0}\right) \\ $$$${cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\:{and}\:\:{cos}\left(\mathrm{2}{x}\right)\sim\:\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow\sqrt{{cos}\left(\mathrm{2}{x}\right)}\sim\sqrt{\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}{x}^{\mathrm{2}} \right)\:=\mathrm{1}−{x}^{\mathrm{2}} \:\Rightarrow{A}\left({x}\right)\sim\:\frac{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}\right)}{{x}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} \:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{2}}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{2}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\rightarrow\frac{\mathrm{3}}{\mathrm{2}}\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:{A}\left({x}\right)\:=\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$

Commented by Tawa1 last updated on 24/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

2)t=θ−(π/6) lim_(t→0) ((sin(t))/((√3) −2cos(t+(π/6)))) =lim_(t→0) ((sint)/((√3) −2(cost.((√3)/2)−sint.(1/2)))) =lim_(t→0) ((sint)/((√3) −(√3) cost+sint)) =lim_(t→0) ((2sin(t/2)cos(t/2))/((√3) ×2sin^2 (t/2)+2sin(t/2)cos(t/2))) =lim_(t→0) ((cos(t/2))/((√3) sin(t/2)+cos(t/2)))=1 or method lim_(θ→((π )/6)) ((sin(θ−(π/6)))/(2[((√3)/2)−cos(θ)])) [((√3)/2)=cos(π/6)] (1/2)lim_(θ→((π )/6)) ((2sin(((θ−(π/6))/2))cos(((θ−(π/6))/2)))/(2sin((((π/6)+θ)/2))sin(((θ−(π/6))/2)))) =(1/2)×(1/(1/2))=1

$$\left.\mathrm{2}\right){t}=\theta−\frac{\pi}{\mathrm{6}} \\ $$$${li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{sin}\left({t}\right)}{\sqrt{\mathrm{3}}\:−\mathrm{2}{cos}\left({t}+\frac{\pi}{\mathrm{6}}\right)} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{sint}}{\sqrt{\mathrm{3}}\:−\mathrm{2}\left({cost}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sint}.\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{sint}}{\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{3}}\:{cost}+{sint}} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{cos}\frac{{t}}{\mathrm{2}}}{\sqrt{\mathrm{3}}\:×\mathrm{2}{sin}^{\mathrm{2}} \frac{{t}}{\mathrm{2}}+\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{cos}\frac{{t}}{\mathrm{2}}} \\ $$$$={li}\underset{{t}\rightarrow\mathrm{0}} {{m}}\:\frac{{cos}\frac{{t}}{\mathrm{2}}}{\sqrt{\mathrm{3}}\:{sin}\frac{{t}}{\mathrm{2}}+{cos}\frac{{t}}{\mathrm{2}}}=\mathrm{1} \\ $$$${or}\:{method} \\ $$$${li}\underset{\theta\rightarrow\frac{\pi\:}{\mathrm{6}}} {{m}}\:\:\frac{{sin}\left(\theta−\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2}\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{cos}\left(\theta\right)\right]}\:\:\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={cos}\frac{\pi}{\mathrm{6}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{li}\underset{\theta\rightarrow\frac{\pi\:}{\mathrm{6}}} {{m}}\:\frac{\mathrm{2}{sin}\left(\frac{\theta−\frac{\pi}{\mathrm{6}}}{\mathrm{2}}\right){cos}\left(\frac{\theta−\frac{\pi}{\mathrm{6}}}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\frac{\pi}{\mathrm{6}}+\theta}{\mathrm{2}}\right){sin}\left(\frac{\theta−\frac{\pi}{\mathrm{6}}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{1} \\ $$

Commented by Tawa1 last updated on 23/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

most welcome...

$${most}\:{welcome}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

3)lim_(x→0) ((1−cosx(√(cos2x)))/x^2 ) lim_(x→0) ((1−cos^2 x(((1−tan^2 x)/(1+tan^2 x))))/(x^2 (1+cosx(√(cos2x)) ))) =lim_(x→0) ((sec^2 x−cos^2 x+sin^2 x)/(sec^2 x×x^2 ×(1+cosx(√(cos2x)) ))) =(1/2)lim_(x→0) (((1/(cos^2 x))−cos^2 x+sin^2 x)/x^2 ) (1/2)lim_(x→0) ((sin^2 x.(1+cos^2 x)+sin^2 x)/(x^2 cos^2 x)) =(1/2)lim_(x→0) (((sinx)/x))^2 ×(((2+cos^2 x)/(cos^2 x))) =(1/2)×1×(3/2)=(3/4)

$$\left.\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}−{cosx}\sqrt{{cos}\mathrm{2}{x}}}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}\left(\frac{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{cosx}\sqrt{{cos}\mathrm{2}{x}}\:\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sec}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{sec}^{\mathrm{2}} {x}×{x}^{\mathrm{2}} ×\left(\mathrm{1}+{cosx}\sqrt{{cos}\mathrm{2}{x}}\:\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}−{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sin}^{\mathrm{2}} {x}.\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)+{sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{{sinx}}{{x}}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{2}+{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by Tawa1 last updated on 23/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

1)lim_(y→0) ((x{sec(x+y)−secx}+ysec(x+y))/y) =x[lim_(y→0) ((sec(x+y)−secx)/y)]+secx =x[lim_(y→0) ((cosx−cos(x+y))/(ycosxcos(x+y)))]+secx =x[lim_(y→0) ((2sin(x+(y/2))sin((y/2)))/(cosxcos(x+y)×(y/2)×2))]+secx =x×((sinx)/(cos^2 x))+secx =xtanxsecx+secx

$$\left.\mathrm{1}\right)\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left\{{sec}\left({x}+{y}\right)−{secx}\right\}+{ysec}\left({x}+{y}\right)}{{y}} \\ $$$$={x}\left[\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{sec}\left({x}+{y}\right)−{secx}}{{y}}\right]+{secx} \\ $$$$={x}\left[\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{cosx}−{cos}\left({x}+{y}\right)}{{ycosxcos}\left({x}+{y}\right)}\right]+{secx} \\ $$$$={x}\left[\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{sin}\left({x}+\frac{{y}}{\mathrm{2}}\right){sin}\left(\frac{{y}}{\mathrm{2}}\right)}{{cosxcos}\left({x}+{y}\right)×\frac{{y}}{\mathrm{2}}×\mathrm{2}}\right]+{secx} \\ $$$$={x}×\frac{{sinx}}{{cos}^{\mathrm{2}} {x}}+{secx} \\ $$$$={xtanxsecx}+{secx} \\ $$

Commented by Tawa1 last updated on 23/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

most welcome...

$${most}\:{welcome}... \\ $$

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