How to show $\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$ I need to prove the result without using L'Hopitals rule
$$\lim\limits_{x \to \infty}[x(\sqrt {x^2+a} - \sqrt {x^2+b})]=\frac{a-b}{2}$$
but this seems quite miraculous to me and I'm not quite sure what to do as everything I do seems to make it worse. Any help would be appreciated.
 A: HINT
Multiply by a special kind of $1$  :
$$\lim\limits_{x \to \infty}\left[x(\sqrt {x^2+a} - \sqrt {x^2+b})\right]=\lim\limits_{x \to \infty}\left[ \dfrac{x(\sqrt {x^2+a} - \sqrt {x^2+b})(\sqrt {x^2+a} + \sqrt {x^2+b})}{\sqrt {x^2+a} + \sqrt {x^2+b}}\right] $$
and use difference of squares
A: A common trick: multiply by $1$.
That is: multiply $x(\sqrt{x^2+a^2}-\sqrt{x^2+b^2})$ by $\frac{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}{x(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$
A: You have another solution using Taylor series. Rewrite $$x\Big(\sqrt {x^2+a} - \sqrt {x^2+b}\Big)=x^2\Big(\sqrt{1+a/x^2}-\sqrt{1+b/x^2}\Big)$$ and remember that, when $y$ is small $\sqrt{1+y}\simeq 1+y/2$. Replace successively $y$ by $a/x^2$ and $b/x^2$ and you are done.
If you continue in the same spirit you coulod show a beautiful thing $$x\Big(\sqrt {x^2+a} - \sqrt {x^2+b}\Big)=\frac{a-b}{2}-\frac{a^2-b^2}{8 x^2}+\frac{a^3-b^3}{16 x^4}-\frac{5
   \left(a^4-b^4\right)}{128 x^6}+O\left(\left(\frac{1}{x}\right)^7\right)$$
A: $\bf{My\; Solution::}$ Given $\displaystyle \lim_{x\rightarrow \infty}\left[x\left(\sqrt{x^2+a}-\sqrt{x^2+b}\right)\right]=\lim_{x\rightarrow \infty}\left[x\cdot x\left\{\left(1+\frac{a}{x^2}\right)^{\frac{1}{2}}-\left(1+\frac{b}{x^2}\right)^{\frac{1}{2}}\right\}\right]$
Now Using Binomial Theorem For Fractional Index....
So $\displaystyle \lim_{x\rightarrow \infty}x^2\left\{\left(1+\frac{a}{2x^2}+......\right)-\left(1+\frac{b}{2x^2}+......\right)\right\} = \left(\frac{a-b}{2}\right)$
