Finding $\lim_{x\to a} \frac{1}{(a^2-x^2)^2}(\frac{a^2+x^2}{ax}-2\sin( \frac{a\pi}{2})\sin(\frac{x\pi}{2}))$ 
Without using the L'Hopital Rule, evaluate the following limit.
  $$\lim_{x\to a} \frac{1}{(a^2-x^2)^2}\left(\dfrac{a^2+x^2}{ax}-2\sin\left( \frac{a\pi}{2}\right)\sin\left(\frac{x\pi}{2}\right) \right)$$
  Where $a$ is an odd integer.
Answer: $\dfrac{\pi^2a^2+4}{16a^4}$

My attempt:
I considered the substitution $x=\lim_\limits{h\to 0}a\cos h$ and got the limit to be:
$$\lim_\limits{h\to 0}\frac{1}{a^4h^4}\left(\frac{1+\cos^2 h}{\cos h}-2\cos^2(ah^2\pi)\right)$$
Aaaand I'm stuck. Please nudge me in the right direction. Profuse thanks. 
 A: I shall focus on the key term
$$y=\frac{1+\cos^2 (h)}{\cos (h)}-2\cos^2(ah^2\pi)$$ and only use
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ Squaring
$$\cos^2(x)=1-x^2+\frac{x^4}{3}-\frac{2 x^6}{45}+O\left(x^8\right)$$
For the first piece
$$\frac{1+\cos^2 (h)}{\cos (h)}=\frac{2-h^2+\frac{h^4}{3}-\frac{2 h^6}{45}+O\left(h^8\right) } {1-\frac{h^2}{2}+\frac{h^4}{24}-\frac{h^6}{720}+O\left(h^8\right) }$$  Use the long division to get
$$\frac{1+\cos^2 (h)}{\cos (h)}=2+\frac{h^4}{4}+\frac{h^6}{12}+O\left(h^8\right)$$ Now, work the second term replacing $x$ by $ah^2\pi$ and ... finish !
A: Let, $x-a=z$.
So, the limit changes to 
$\lim \limits_{z \to 0} \frac{\frac{a^2+x^2}{ax}-2\text{sin}(\frac{aπ}{2})\text{sin}(\frac{xπ}{2})}{(a^2-x^2)^2}$
$=\lim \limits_{z \to 0} \frac{\frac{(x-a)^2}{ax}+2-2\text{sin}(\frac{aπ}{2})\text{sin}(\frac{xπ}{2})}{(z(x+a))^2} $
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{\frac{z^2}{a^2}+1-\text{cos}(\frac{z\pi}{2})+1+\text{cos}(aπ+\frac{z\pi}{2})}{z^2})$
(Because $x+a=2a+z$)
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{1}{a^2}+\frac{1-\text{cos}(\frac{zπ}{2})+1-\text{cos}(\frac{zπ}{2})}{z^2})$
(as $a$ is odd, $\text{cos}(aπ+\frac{zπ}{2}))=-\text{cos}(\frac{zπ}{2})$)
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{1}{a^2}+\frac{4\text{sin}^2(\frac{zπ}{4})}{z^2})$
(For odd $a$, $\text{cos}(\frac{aπ}{2})=0$)
$=\frac{1}{4a^2}(\lim \limits_{z \to 0} \frac{1}{a^2}+\frac{π^2\text{sin}^2(\frac{zπ}{4})}{4(\frac{zπ}{4})^2})$
$=\frac{(πa)^2+4}{16a^4}$ .... proved .
