Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ended up with
$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
Which I'm not sure what to do with.

We may suppose $x>0$, so
\begin{align}\sqrt{x^2+ax}+\sqrt{x^2+bx}&=x\biggl(\sqrt{1+\frac ax}+\sqrt{1+\frac bx}   \,\biggr)=x\biggl(1+\frac a{2x}+1+\frac b{2x} +o\Bigl(\frac1x\Bigr) \biggr)\\
&=2x+\frac{a+b}2+o(1),\end{align}
so that the denominator is equivalent at $\infty$ to $2x$, and
$$\sqrt{x^2 +ax} - \sqrt{x^2 +bx}=\frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}\sim_\infty\frac{(a-b)x}{2x}=\frac{a-b}{2}.$$
