<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] $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}
=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\sqrt{x^2 +ax} - \sqrt{x^2 +bx}}=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\vert{x}\vert\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\right)}=\frac{(a-b)}{2}\lim_{x\rightarrow \infty}\frac{x}{\vert x\vert}=\frac{a-b}{2}.$$
Using that $\vert x\vert=\sqrt{x^{2}}$.
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