Evaluate $\lim_{x\ \to\ \infty}\left(\,\sqrt{\,x^{4} + ax^{2} + 1\,}\, - \,\sqrt{\,x^{4} + bx^{2} +1\,}\,\right)$

Question

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 $\displaystyle{{1 \over 2}\left(\,a - b\,\right)}$.

Answer 1

Hint:

$$\lim_{x \to \infty} \frac{(\sqrt{x^4 + ax^2 +1} - \sqrt{x^4 + bx^2 +1})(\sqrt{x^4 + ax^2 +1} + \sqrt{x^4 + bx^2 +1})}{\sqrt{x^4 + ax^2 +1} + \sqrt{x^4 + bx^2 +1}}$$

The correct answer is $\frac{1}{2}(a-b)$.

Answer 2

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 $$x^2\Big(\sqrt{1 + \dfrac{a}{x^2} +\dfrac{1}{x^4}} - \sqrt{1 + \dfrac{b}{x^2} +\dfrac{1}{x^4}}\Big)=\frac{a-b}{2}+\frac{b^2-a^2}{8 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$

Answer 3

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 $x$.

Answer 4

Substitute $x=1/t,$ then our expression becomes $$\lim_{t \to 0} \frac{\sqrt{1+at^2+t^4}-\sqrt{1+bt^2+t^4}}{t^2}$$ Note that here we can apply the L'Hospitals rule. $$\frac{\frac{1}{2}(1+at^2+t^4)^{-1/2}(2at+4t^3)}{2t} \to \frac {a}{2}$$ as $t \to 0.$
Therefore it easy to see that required limit is $$\frac{a-b}{2}.$$