<s>[INST] 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. 
 [/INST] 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 !
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