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Question Number 194335 by mathlove last updated on 04/Jul/23

Answered by cortano12 last updated on 04/Jul/23

let x−1= u lim_(u→0) ((1−sin ((π/(u^2 −u+2))))/u^2 ) = lim_(u→0) ((1−cos ((π/2)−(π/(u^2 −u+2))))/u^2 ) = lim_(u→0) ((1−cos π(((u^2 −u)/(2(u^2 −u+2)))))/u^2 ) = lim_(u→0) (((2sin^2 (((πu(u−1))/(4(u^2 −u+2)))))/u^2 )) = ((2.π^2 .1)/(4^2 .2^2 )) = determinant (((π^2 /(32))))

$$\:\: \\ $$$$ \mathrm{let}\:\mathrm{x}−\mathrm{1}=\:\mathrm{u} \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{u}^{\mathrm{2}} −\mathrm{u}+\mathrm{2}}\right)}{\mathrm{u}^{\mathrm{2}} } \\ $$$$\:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{u}^{\mathrm{2}} −\mathrm{u}+\mathrm{2}}\right)}{\mathrm{u}^{\mathrm{2}} } \\ $$$$\:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\pi\left(\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{u}}{\mathrm{2}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{u}+\mathrm{2}\right)}\right)}{\mathrm{u}^{\mathrm{2}} } \\ $$$$\:=\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\pi\mathrm{u}\left(\mathrm{u}−\mathrm{1}\right)}{\mathrm{4}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{u}+\mathrm{2}\right)}\right)}{\mathrm{u}^{\mathrm{2}} }\right) \\ $$$$\:=\:\frac{\mathrm{2}.\pi^{\mathrm{2}} .\mathrm{1}}{\mathrm{4}^{\mathrm{2}} .\mathrm{2}^{\mathrm{2}} }\:=\:\begin{array}{|c|}{\frac{\pi^{\mathrm{2}} }{\mathrm{32}}}\\\hline\end{array} \\ $$$$\: \\ $$

Commented by mathlove last updated on 04/Jul/23

tnk

$${tnk} \\ $$

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