Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 82485 by naka3546 last updated on 21/Feb/20

Commented by abdomathmax last updated on 21/Feb/20

let f(x)=(6/t^2 )−((sin(6t))/(t^3 cos^2 (3t))) ⇒ f(x)=(1/t^2 ){6−((sin(6t))/(t cos^2 (3t)))} we have sin(6t)∼6t cos^2 (3t)=((1+cos(6t))/2) ∼(1/2) +(1/2)(1−((36t^2 )/2)) =(1/2)+(1/2)−9t^2 =1−9t^2 ⇒ f(x)∼(1/t^2 ){6−((6t)/(t(1−9t^2 )))} =(1/t^2 ){6−(6/(1−9t^2 ))} =(6/t^2 )×((1−9t^2 −1)/((1−9t^2 ))) =((−54)/(1−9t^2 )) →−54 ⇒ lim_(x→0) f(x)=−54

$${let}\:{f}\left({x}\right)=\frac{\mathrm{6}}{{t}^{\mathrm{2}} }−\frac{{sin}\left(\mathrm{6}{t}\right)}{{t}^{\mathrm{3}} {cos}^{\mathrm{2}} \left(\mathrm{3}{t}\right)}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left\{\mathrm{6}−\frac{{sin}\left(\mathrm{6}{t}\right)}{{t}\:{cos}^{\mathrm{2}} \left(\mathrm{3}{t}\right)}\right\}\:{we}\:{have}\: \\ $$$${sin}\left(\mathrm{6}{t}\right)\sim\mathrm{6}{t} \\ $$$${cos}^{\mathrm{2}} \left(\mathrm{3}{t}\right)=\frac{\mathrm{1}+{cos}\left(\mathrm{6}{t}\right)}{\mathrm{2}}\:\sim\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{36}{t}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{9}{t}^{\mathrm{2}} \:=\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left\{\mathrm{6}−\frac{\mathrm{6}{t}}{{t}\left(\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} \right)}\right\}\:=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left\{\mathrm{6}−\frac{\mathrm{6}}{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{6}}{{t}^{\mathrm{2}} }×\frac{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} \right)}\:=\frac{−\mathrm{54}}{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} }\:\rightarrow−\mathrm{54}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)=−\mathrm{54} \\ $$

Commented by jagoll last updated on 22/Feb/20

lim_(t→0) ((6t cos^2 t −sin 6t)/(t^3 cos^2 t)) = lim_(t→0) (1/(cos^2 t)) × lim_(t→0) ((6cos^2 t−6tsin 2t−6cos 6t)/(3t^2 )) = 1 × lim_(t→0) ((−6sin 2t−(6sin 2t+12tcos 2t)+36sin 6t)/(6t)) = lim_(t→0) ((−24cos 2t−(12cos 2t−24tsin 2t)+216cos 6t)/6) = ((−36+216)/6) = −6+ 36 = 30?

$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}{t}\:\mathrm{cos}\:^{\mathrm{2}} {t}\:−\mathrm{sin}\:\mathrm{6}{t}}{{t}^{\mathrm{3}} \:\mathrm{cos}\:^{\mathrm{2}} {t}}\:=\: \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {t}}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6cos}\:^{\mathrm{2}} {t}−\mathrm{6}{t}\mathrm{sin}\:\mathrm{2}{t}−\mathrm{6cos}\:\mathrm{6}{t}}{\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{1}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{6sin}\:\mathrm{2}{t}−\left(\mathrm{6sin}\:\mathrm{2}{t}+\mathrm{12}{t}\mathrm{cos}\:\mathrm{2}{t}\right)+\mathrm{36sin}\:\mathrm{6}{t}}{\mathrm{6}{t}} \\ $$$$=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{24cos}\:\mathrm{2}{t}−\left(\mathrm{12cos}\:\mathrm{2}{t}−\mathrm{24}{t}\mathrm{sin}\:\mathrm{2}{t}\right)+\mathrm{216cos}\:\mathrm{6}{t}}{\mathrm{6}} \\ $$$$=\:\frac{−\mathrm{36}+\mathrm{216}}{\mathrm{6}}\:=\:−\mathrm{6}+\:\mathrm{36}\:=\:\mathrm{30}? \\ $$

Commented by naka3546 last updated on 22/Feb/20

cos t or cos 3t ?

$$\mathrm{cos}\:{t}\:\:{or}\:\:\mathrm{cos}\:\mathrm{3}{t}\:? \\ $$

Commented by john santu last updated on 22/Feb/20

this answer is −18

$${this}\:{answer}\:{is}\:−\mathrm{18} \\ $$

Answered by john santu last updated on 22/Feb/20

lim_(t→0) (6/t^2 ) − ((2sin 3t cos 3t)/(t^3 cos^2 3t)) = lim_(t→0) (6/t^2 ) − ((2tan 3t)/t^3 ) = lim_(t→0) ((6t−2tan 3t)/t^3 ) = lim_(t→0) ((6−6sec^2 3t)/(3t^2 )) = lim_(t→0) ((2(1−(1/(cos^2 3t))))/t^2 ) =2× lim_(t→0 ) ((cos^2 3t−1)/(t^2 cos^2 3t)) = 2×lim_(t→0) (1/(cos^2 3t)) × lim_(t→0) ((−sin^2 3t)/t^2 ) = 2 × 1 × (−9) = −18

$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}}{{t}^{\mathrm{2}} }\:−\:\frac{\mathrm{2sin}\:\mathrm{3}{t}\:\mathrm{cos}\:\mathrm{3}{t}}{{t}^{\mathrm{3}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}{t}}\:= \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}}{{t}^{\mathrm{2}} }\:−\:\frac{\mathrm{2tan}\:\mathrm{3}{t}}{{t}^{\mathrm{3}} }\:=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}{t}−\mathrm{2tan}\:\mathrm{3}{t}}{{t}^{\mathrm{3}} } \\ $$$$=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{6}−\mathrm{6sec}\:^{\mathrm{2}} \mathrm{3}{t}}{\mathrm{3}{t}^{\mathrm{2}} }\:=\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}{t}}\right)}{{t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}×\:\underset{{t}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}{t}−\mathrm{1}}{{t}^{\mathrm{2}} \:\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}{t}}\:=\:\mathrm{2}×\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}{t}}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{t}}{{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:×\:\mathrm{1}\:×\:\left(−\mathrm{9}\right)\:=\:−\mathrm{18} \\ $$

Answered by Henri Boucatchou last updated on 22/Feb/20

At V(0), sin6t∼6t, (6/t^2 )−((sin6t)/(t^3 (1−sin^2 3t)))∼(6/t^2 )(1−(1/(1−(3t)^2 ))) ∼(6/t^2 )(((1−9t^2 −1)/(1−9t^2 )))∼−54(1/(1−9t^2 )) ⇒ lim_(x→0) ((6/t^2 )−((sin6t)/(t^3 cos^2 t)))=lim_(x→0) (−54(1/(1−9t^2 )))=−54

$${At}\:\:{V}\left(\mathrm{0}\right),\:\:{sin}\mathrm{6}{t}\sim\mathrm{6}{t},\:\frac{\mathrm{6}}{{t}^{\mathrm{2}} }−\frac{{sin}\mathrm{6}{t}}{{t}^{\mathrm{3}} \left(\mathrm{1}−{sin}^{\mathrm{2}} \mathrm{3}{t}\right)}\sim\frac{\mathrm{6}}{{t}^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\left(\mathrm{3}{t}\right)^{\mathrm{2}} }\right) \\ $$$$\sim\frac{\mathrm{6}}{{t}^{\mathrm{2}} }\left(\frac{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} }\right)\sim−\mathrm{54}\frac{\mathrm{1}}{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{6}}{{t}^{\mathrm{2}} }−\frac{{sin}\mathrm{6}{t}}{{t}^{\mathrm{3}} {cos}^{\mathrm{2}} {t}}\right)=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(−\mathrm{54}\frac{\mathrm{1}}{\mathrm{1}−\mathrm{9}{t}^{\mathrm{2}} }\right)=−\mathrm{54} \\ $$

Commented by john santu last updated on 22/Feb/20

not correct lim_(x→0) 1−sin^2 3t ∼ 1−(3t)^2

$${not}\:{correct}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{3}{t}\:\sim\:\mathrm{1}−\left(\mathrm{3}{t}\right)^{\mathrm{2}} \\ $$

Commented by Henri Boucatchou last updated on 22/Feb/20

Sorry, take t=x

$${Sorry},\:{take}\:{t}={x} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com