Least value of $L$ for $\sqrt{x^2+ax}-\sqrt{x^2+bx}0$ 
If $a$ and $b$ are positive real numbers such that $a-b=2$ , then the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$.

This question is similar to this one, but I don't want to apply the concept of limit. I want to use application of derivative in solving this problem.
I thought of putting $\sqrt{x^2+ax}-\sqrt{x^2+bx}=L$ and then using $\frac{dL}{dx}=0$, but the equation is coming in the form of $a+b$ which I am not able to solve.

We have $$\sqrt{x^2+ax}-\sqrt{x^2+bx}=\frac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$$
for all $x>0$.  Because $\sqrt{1+\frac{a}{x}}>1$ and $\sqrt{1+\frac{b}{x}}>1$ for all $x>0$, we get
$$\sqrt{x^2+ax}-\sqrt{x^2+bx}< \frac{a-b}{2}.$$
This shows that $L\leq \frac{a-b}{2}$.  As
$$\lim_{x\to\infty}\left(\sqrt{x^2+ax}-\sqrt{x^2+bx}\right)=\lim_{x\to\infty}\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{a-b}{2},$$
we must have $L=\frac{a-b}{2}$.  (By the way, you cannot avoid using limits in this problem because $L$ is attained as $x$ goes to infinity.  Taking derivative of $\sqrt{x^2+ax}-\sqrt{x^2+bx}$ will not give you a maximizing point on $(0,\infty)$ because the maximum does not exist.)
