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Question Number 168143 by cortano1 last updated on 04/Apr/22

               lim_(x→0)  ((1−cos (1−cos x))/x^4 ) =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{4}} }\:=?\:\:\:\:\:\: \\ $$

Answered by qaz last updated on 04/Apr/22

  lim_(x→0) ((1−cos (1−cos x))/x^4 )=lim_(x→0) ((1−cos ((1/2)x^2 +...))/x^4 )  =lim_(x→0) (((1/2)((1/2)x^2 +...)^2 )/x^4 )=lim_(x→0) (((1/8)x^4 +o(x^4 ))/x^4 )=(1/8)

$$ \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} }=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +...\right)}{\mathrm{x}^{\mathrm{4}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} +...\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{4}} }=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{8}}\mathrm{x}^{\mathrm{4}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{4}} \right)}{\mathrm{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Answered by Mathspace last updated on 04/Apr/22

1−cosx∼(x^2 /2) ⇒cos(1−cosx)∼cos((x^2 /2))  ∼1−(x^4 /8) ⇒1−cos(1−cosx)∼(x^4 /8)  ⇒((1−cos(1−cosx))/x^4 )∼(1/8) ⇒  lim_(x→0) ((1−cos(1−cosx))/x^4 )=(1/8)

$$\mathrm{1}−{cosx}\sim\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{cos}\left(\mathrm{1}−{cosx}\right)\sim{cos}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\sim\mathrm{1}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}}\:\Rightarrow\mathrm{1}−{cos}\left(\mathrm{1}−{cosx}\right)\sim\frac{{x}^{\mathrm{4}} }{\mathrm{8}} \\ $$$$\Rightarrow\frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cosx}\right)}{{x}^{\mathrm{4}} }\sim\frac{\mathrm{1}}{\mathrm{8}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cosx}\right)}{{x}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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