Menu Close

lim-x-0-sin-4-picos-x-1-cos-1-cos-1-cos-x-

Tinku Tara 0 Comments
Pin




Question Number 192625 by Subhi last updated on 23/May/23
lim_(x→0) ((sin^4 (πcos(x)))/(1−cos(1−cos(1−cos(x)))))
$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}^{\mathrm{4}} \left(\pi{cos}\left({x}\right)\right)}{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)} \\ $$
Commented by Subhi last updated on 23/May/23
I got 8π^4
$${I}\:{got}\:\mathrm{8}\pi^{\mathrm{4}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 23/May/23
L=lim_(x→0) ((sin^4 (πcosx))/(1−cos(1−cos(1−cosx)))) =lim_(x→0) ((sin^4 (π−((πx^2 )/2)))/(1−(1−(x^8 /(128))))) [sint ∼_(t→0) t; cost ∼_(t→0) 1−(t^2 /2)] =lim_(x→0) ((sin^4 (((πx^2 )/2)))/(x^8 /(128)))=lim_(x→0) (((π^4 x^8 )/(16))/(x^8 /(128)))= determinant (((8π^4 )))
$$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{cos}{x}\right)}{\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}{x}\right)\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{8}} }{\mathrm{128}}\right)}\:\left[\mathrm{sin}{t}\:\underset{{t}\rightarrow\mathrm{0}} {\sim}\:{t};\:\mathrm{cos}{t}\:\underset{{t}\rightarrow\mathrm{0}} {\sim}\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\frac{{x}^{\mathrm{8}} }{\mathrm{128}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\pi^{\mathrm{4}} {x}^{\mathrm{8}} }{\mathrm{16}}}{\frac{{x}^{\mathrm{8}} }{\mathrm{128}}}=\begin{array}{|c|}{\mathrm{8}\pi^{\mathrm{4}} }\\\hline\end{array} \\ $$
Commented by Subhi last updated on 23/May/23
thanks boss I solved it by that way↓↓
$${thanks}\:{boss} \\ $$$${I}\:{solved}\:{it}\:{by}\:{that}\:{way}\downarrow\downarrow \\ $$
Answered by Subhi last updated on 23/May/23
lim_(x→0) ((sin^4 (πcos(x)))/(1−cos(1−cos(1−cos(x))))) sin(π−θ) = sin(θ) x→0 ⇛ cos(x)→1 ⇛ π−πcos(x)→0 lim_(π−πcos(x)→0) ((sin^4 (π−πcos(x)).(π−πcos(x))^4 )/((π−πcos(x))^4 .(1−cos(1−cos(1−cos(x))))) note that lim_(x→0) ((sin(x))/x)=1 lim_(x→0) (((π−πcos(x))^4 )/(1−cos(1−cos(1−cos(x)))) 1−cos(x)=2sin^2 ((x/2)) 1−cos(1−cos(1−cos(x))=1−cos(1−cos(2sin^2 ((x/2))) =1−cos(2sin^2 (sin^2 ((x/2)))=2sin^2 (sin^2 (sin^2 ((x/2))) lim_(x→0) (((π−πcos(x))^4 )/(2sin^2 (sin^2 (sin^2 ((x/2)))))) x→0 ⇛ (x/2)→0 ⇛sin^2 (x/2)→0⇛sin^2 (sin^2 ((x/2)))→0 lim_(sin^2 (sin^2 ((x/2)))→0) (((π−πcos(x))^4 .(sin^2 (sin^2 ((x/2)))^2 )/(2sin^2 (sin^2 (sin^2 ((x/2)))).sin^4 (sin^2 ((x/2))))) lim_(x→0) (((π−πcos(x))^4 )/(2sin^4 (sin^2 ((x/2))))) lim_(sin^2 ((x/2))→0) (((π−πcos(x))^4 .(sin^8 ((x/2))))/(2sin^4 (sin^2 ((x/2))).sin^8 ((x/2)))) lim_(x→0) ((π^4 (1−cos(x))^4 )/(2sin^8 ((x/2)))).(8/8) 1−cos(x)=2sin^2 ((x/2)) (1−cos(x))^4 =16sin^8 ((x/2)) answer = 8π^4
$$ \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}^{\mathrm{4}} \left(\pi{cos}\left({x}\right)\right)}{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)} \\ $$$${sin}\left(\pi−\theta\right)\:=\:{sin}\left(\theta\right) \\ $$$${x}\rightarrow\mathrm{0}\:\:\:\:\:\:\:\Rrightarrow\:{cos}\left({x}\right)\rightarrow\mathrm{1}\:\:\:\:\Rrightarrow\:\:\pi−\pi{cos}\left({x}\right)\rightarrow\mathrm{0} \\ $$$${lim}_{\pi−\pi{cos}\left({x}\right)\rightarrow\mathrm{0}} \frac{{sin}^{\mathrm{4}} \left(\pi−\pi{cos}\left({x}\right)\right).\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} }{\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} .\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)\right.} \\ $$$${note}\:{that}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}\left({x}\right)}{{x}}=\mathrm{1} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} }{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right.} \\ $$$$\mathrm{1}−{cos}\left({x}\right)=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)=\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\right.\right. \\ $$$$=\mathrm{1}−{cos}\left(\mathrm{2}{sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)=\mathrm{2}{sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\right.\right. \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} }{\mathrm{2}{sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\right)} \\ $$$${x}\rightarrow\mathrm{0}\:\:\Rrightarrow\:\frac{{x}}{\mathrm{2}}\rightarrow\mathrm{0}\:\:\Rrightarrow{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\rightarrow\mathrm{0}\Rrightarrow{sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\rightarrow\mathrm{0} \\ $$$${lim}_{{sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\rightarrow\mathrm{0}} \frac{\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} .\left({sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} \right.}{\mathrm{2}{sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\right).{sin}^{\mathrm{4}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} }{\mathrm{2}{sin}^{\mathrm{4}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)} \\ $$$${lim}_{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\rightarrow\mathrm{0}} \frac{\left(\pi−\pi{cos}\left({x}\right)\right)^{\mathrm{4}} .\left({sin}^{\mathrm{8}} \left(\frac{{x}}{\mathrm{2}}\right)\right)}{\mathrm{2}{sin}^{\mathrm{4}} \left({sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right).{sin}^{\mathrm{8}} \left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\pi^{\mathrm{4}} \left(\mathrm{1}−{cos}\left({x}\right)\right)^{\mathrm{4}} }{\mathrm{2}{sin}^{\mathrm{8}} \left(\frac{{x}}{\mathrm{2}}\right)}.\frac{\mathrm{8}}{\mathrm{8}} \\ $$$$\mathrm{1}−{cos}\left({x}\right)=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{1}−{cos}\left({x}\right)\right)^{\mathrm{4}} =\mathrm{16}{sin}^{\mathrm{8}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$$${answer}\:=\:\mathrm{8}\pi^{\mathrm{4}} \\ $$$$ \\ $$
Answered by MM42 last updated on 24/May/23
Another solution π−πcosx=u ⇒ 1−cosx=(u/π) ⇒L=lim_(u→0) ((sin^4 (π−u))/(1−cos(1−cos((u/π))))) = lim_(u→0) ((sin^4 (u))/(1−cos((u^2 /(2π^2 ))))) = lim_(u→0) (u^4 /(u^4 /(8π^2 ))) = 8π^2 ✓
$${Another}\:{solution} \\ $$$$\pi−\pi{cosx}={u}\:\Rightarrow\:\mathrm{1}−{cosx}=\frac{{u}}{\pi} \\ $$$$\Rightarrow{L}={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{{sin}^{\mathrm{4}} \left(\pi−{u}\right)}{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\frac{{u}}{\pi}\right)\right)}\:= \\ $$$${lim}_{{u}\rightarrow\mathrm{0}} \:\frac{{sin}^{\mathrm{4}} \left({u}\right)}{\mathrm{1}−{cos}\left(\frac{{u}^{\mathrm{2}} }{\mathrm{2}\pi^{\mathrm{2}} }\right)}\:= \\ $$$${lim}_{{u}\rightarrow\mathrm{0}} \:\:\frac{{u}^{\mathrm{4}} }{\frac{{u}^{\mathrm{4}} }{\mathrm{8}\pi^{\mathrm{2}} }}\:=\:\mathrm{8}\pi^{\mathrm{2}} \:\:\checkmark \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *