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Question Number 32407 by saru53424@gmail.com last updated on 24/Mar/18

$$\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos \mathbf{x})}{\mathbf{x}\times \mathbf{x}\times \mathbf{x}\times \mathbf{x}}$$

Commented by abdo imad last updated on 24/Mar/18

$$ifyoumean{lim}_{x\to 0}\frac{1-cos(1-cosx)}{x^4}inthiscaseits$$$$bettertousehospitaltheoremletputu\left(x\right)=1-cos(1-cosx)$$$$andv\left(x\right)=x^4\Rightarrow v^{{}^{\prime }}\left(x\right)=4x^3\Rightarrow v^{{}^{″}}\left(x\right)=12x^2$$$$\Rightarrow v^{\left(3\right)}\left(x\right)=24x\Rightarrow v^{\left(4\right)}\left(x\right)=24$$$$u^{{}^{\prime }}\left(x\right)=sinxsin(1-cosx)\Rightarrow u^{{}^{″}}\left(x\right)=cosxsin(1-cosx)$$$$+sinx.sinxcos(1-cosx)=cosxsin(1-cosx)$$$$+{sin}^{2}xcos(1-cosx)$$$$u^{\left(3\right)}\left(x\right)=-sinxsin(1-cosx)+cosx.sinxcos(1-cosx)$$$$+2sinxcosx.cos(1-cosx)-{sin}^{2}xsin(1-cosx)$$$$=-sinxsin(1-cosx)+\frac{1}{2}sin\left(2x\right)cos(1-cosx)$$$$+sin\left(2x\right).cos(1-cosx)-{sin}^{2}xsin(1-cosx)$$$$u^{\left(4\right)}\left(x\right)=-cosxsin(1-cosx)-sinxsinxcos(1-cosx)$$$$+cos\left(2x\right)cos(1-cosx)-\frac{1}{2}sin\left(2x\right)sinxsin(1-cosx)$$$$+2cos\left(2x\right)cos(1-cosx)-sin\left(2x\right)sinxsin(1-cosx)$$$$-2sinxcosxsin(1-cosx)-{sin}^{3}cos(1-cosx)$$$$u^{\left(4\right)}\left(0\right)=1+2=3\Rightarrow {lim}_{x\to 0}\frac{1-cos(1-cosx)}{x^4}=\frac{3}{24}=\frac{1}{8}$$$$+\left(perhaps\right).....$$

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