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Prove that $\lim_{x\to a}\frac{1}{(x^2-a^2)^2}\left(\frac{a^2+x^2}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)=\frac{\pi^2 a^2+4}{16a^4}$ where $a$ is an odd integer.

I tried to apply L Hospital rule in this question but it is not coming in $\frac{0}{0}$ form,neither series expansion seems to be helpful.What should i do to prove this limit.

user1442
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    $(x^2-a^2)^2\to0$ and $\left(\frac{a^2+x^2}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)\to 2-2=0$ – joefu Sep 26 '15 at 14:05
  • Forget L'Hopital, write $x=a(1+u)$ with $u\to0$, note that $(x^2-a^2)^2\sim4a^2u^2$ and expand to order $u^2$ the parenthesis. For example, $$\frac{a^2+x^2}{ax}=\frac{1}{1+u}+1+u=2+u^2+o(u^2),$$ and $\cos(a\pi/2)=0$ hence $$\sin(\pi x/2)=\sin(\pi a/2)+(\pi au/2)\cos(a\pi/2)-(\pi au/2)^2\sin(a\pi/2)/2+o(u^2)$$ is $$\sin(\pi x/2)=\sin(\pi a/2)-(\pi au/2)^2\sin(a\pi/2)/2+o(u^2),$$ which, noting that $\sin(a\pi/2)^2=1$, yields the result. – Did Sep 26 '15 at 14:41

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We have $$\lim_{x\rightarrow a}\frac{1}{\left(x^{2}-a^{2}\right)^{2}}\left(\frac{x^{2}+a^{2}}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)= $$ $$=\frac{1}{4a^{2}}\lim_{x\rightarrow a}\frac{1}{\left(x-a\right)^{2}}\left(\frac{x^{2}+a^{2}}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)=$$ $$=\frac{1}{4a^{4}}\lim_{x\rightarrow a}\frac{x^{2}+a^{2}-2ax\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)}{\left(x-a\right)^{2}} $$ so apply De L'Hopital twice $$=\frac{1}{4a^{4}}\lim_{x\rightarrow a}\frac{2-2a\pi\cos\left(\frac{\pi x}{2}\right)\sin\left(\frac{a\pi}{2}\right)+2^{-1}\pi^{2}ax\sin\left(\frac{\pi x}{2}\right)\sin\left(\frac{a\pi}{2}\right)}{2}=\frac{4+\pi^{2}a^{2}}{16a^{4}}. $$

Marco Cantarini
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