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 could also use the basic Taylor Series expansion $\left( 1 + \frac{1}{t}\right)^{\lambda} = 1+\frac{\lambda}{t} + O(1/t^2)$ as follows:
$$x^2 \Big( \underbrace{ \sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} }_{1+\frac{a}{2x^2} + O(\frac{1}{x^3})} - \underbrace{ \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}} }_{1+\frac{b}{2x^2}+O(\frac{1}{x^3})} \Big) = x^2 \Big( \frac{a-b}{2x^2} + O\big(\frac{1}{x^3}\big) \Big) = \frac{a-b}{2} + O\big( \frac{1}{x} \big)$$
since the extra term $\frac{1}{x^4}$ in each square root is negligble compared to $\frac{a}{x^2}$ and $\frac{b}{x^2}$ for large $x$.
