<s>[INST] 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.
 [/INST] $$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}=
\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} + \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}}=
\lim_{x\rightarrow \infty} \frac{ax-bx}{|x+\frac{a}{2}| + |x+\frac{b}{2}|}
$$
this is 
$$\lim_{x\rightarrow +\infty} \frac{(a-b)x}{2x+\frac{a+b}{2}}=\color{blue}{\frac{a-b}{2}}$$
as $x\rightarrow +\infty$, and
$$\lim_{x\rightarrow -\infty} \frac{(a-b)x}{-2x-\frac{a+b}{2}}=\color{blue}{-\frac{a-b}{2}}$$
as $x\rightarrow -\infty$.

Of course I have done the first from you have been stucked, so you can do the main limit in this way 
$$\lim_{x\rightarrow \infty} \sqrt{(x+\frac{a}{2})^2 -\frac{a^2}{4}} - \sqrt{(x+\frac{b}{2})^2 -\frac{b^2}{4}}=\lim_{x\rightarrow \infty}|x+\frac{a}{2}| - |x+\frac{b}{2}|$$
and obtain the same results.
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