Finding limx→a1(a2−x2)2(a2+x2ax−2sin⁡(aπ2)sin⁡(xπ2)) My attempt: I considered the substitution x=limh→0acos⁡h and got the limit to be: limh→01a4h4(1+cos2hcos⁡h−2cos2(ah2π))

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Finding limx→a1(a2−x2)2(a2+x2ax−2sin⁡(aπ2)sin⁡(xπ2))  My attempt:  I considered the substitution x=limh→0acos⁡h and got the limit to be:  limh→01a4h4(1+cos2hcos⁡h−2cos2(ah2π))

Fasaniu · Accepted Answer

I shall focus on the key term  y=1+cos2hcos⁡h−2cos2(ah2π)  and only use   cos⁡(x)=1−x22+x424−x6720+O(x8)   Squaring   cos2(x)=1−x2+x43−2x645+O(x8)   For the first piece   1+cos2hcos⁡h=2−h2+h43−2h645+O(h8)1−h22+h424−h6720+O(h8)   Use the long division to get   1+cos2hcos⁡h=2+h44+h612+O(h8)   Now, work the second term replacing x by ah2π

movingsupplyw1 · Accepted Answer

Let, x-a=z   So, the limit changes to   limz→0a2+x2ax−2sin⁡(aπ2)sin⁡(xπ2)(a2−x2)2  =limz→0(x−a)2ax+2−2sin⁡(aπ2)sin⁡(xπ2)(z(x+a))2   =14a2(limz→0z2a2+1−cos⁡(zπ2)+1+cos⁡(aπ+zπ2)z2)   (Because x+a=2a+z)   =14a2(limz→01a2+1−cos⁡(zπ2)+1−cos⁡(zπ2)z2)   (as a is odd, cos⁡(aπ+zπ2))=−cos⁡(zπ2))   =14a2(limz→01a2+4sin2(zπ4)z2)  (For odd a, cos⁡(aπ2)=0)  =14a2(limz→01a2+π2sin2(zπ4)4(zπ4)2)   =(πa)2+416a4 ... proved

Finding \lim_{x \to a} \frac{1}{(a^2-x^2)^2}(\frac{a^2+x^2}{ax}-2 \sin (\frac{a \pi}{2}) \sin (\frac{x

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