Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$ 
I rewrote the function to the form
  $$
x^{2}\left(\,
\sqrt{\,1 + \dfrac{a}{x^{2}} + \dfrac{1}{x^{4}}\,}\,-\,
\sqrt{\,1 + \dfrac{b}{x^{2}} + \dfrac{1}{x^{4}}\,}\,\right)
$$
  and figured that the answer would be $0$, but apparently this is wrong.
The correct answer is $\displaystyle{{1 \over 2}\left(\,a - b\,\right)}$. 


You correctly started with $$x^2\Big(\sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} - \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}}\Big)$$ Now consider that, for small values of $\epsilon $, $$\sqrt{1+\epsilon}=1+\frac{\epsilon }{2}-\frac{\epsilon ^2}{8}+\frac{\epsilon ^3}{16}+O\left(\epsilon
   ^4\right)$$ and replace first $\epsilon$ by $\dfrac{a}{x^2} +\dfrac{1}{x^4}$ and then by $\dfrac{b}{x^2} +\dfrac{1}{x^4}$. You then arrive to $$x^2\Big(\sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} - \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}}\Big)=\frac{a-b}{2}+\frac{b^2-a^2}{8 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$
