Prove that $\lim_{x\to\infty}\left\{\sqrt{ax^2 + bx + c} -\sqrt{px^2 + qx + r}~\right\}=\frac{b-q}{2\sqrt a}$

Question

Prove that $$\lim_{x\to\infty}\left\{\sqrt{ax^2 + bx + c} -\sqrt{px^2 + qx + r}~ \right\} = \frac{b-q}{2\sqrt a}$$

For $ a=p$

Do you know this formula? How to prove this?

Answer 1

When $a=p$,$$\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}=\frac{(ax^2+bx+c)-(px^2+qx+r)}{\sqrt{ax^2+bx+c}+\sqrt{px^2+qx+r}}\sim\frac{(b-q)x}{(\sqrt a+\sqrt p)x}.$$

Answer 2

If you know already derivatives then you can calculate this limit as follows in a very convenient way:

• Set $x = \frac{1}{t}$ and consider $t\to 0^+$:

\begin{eqnarray*}\sqrt{ax^2 + bx + c} -\sqrt{ax^2 + qx + r} & \stackrel{x = \frac{1}{t}}{=} & \frac{\overbrace{\sqrt{a + bt + ct^2}}^{f(t):=} - \overbrace{\sqrt{a + qt + rt^2}}^{g(t):=}}{t} \\ & = & \frac{\sqrt{a + bt + ct^2} -\sqrt{a} + \sqrt{a}- \sqrt{a + qt + rt^2}}{t} \\ & \stackrel{t\to 0^+}{\longrightarrow} & f'(0) -g'(0) \\ & = & \frac{b}{2\sqrt{a}} - \frac{q}{2\sqrt{a}} \end{eqnarray*}