<s>[INST] Limit of difference of two square roots I need to find the limit, not sure what to do.
$\lim_{x \to \infty} \sqrt{x^2 +ax} - \sqrt{x^2 +bx}$
I am pretty sure I have to divide by the largest degree which is x^2 but that gets me some weird numbers that don't seem to help.
 [/INST] This answer may be a bit longer, but I've tried to include every step in the process.
\begin{equation} \label{eq1}
\begin{split}
\lim_{x \to \infty} \sqrt{x^2 + ax} - \sqrt{x^2 + bx} & = \lim_{x \to \infty} \left(\sqrt{x^2 + ax} - \sqrt{x^2 + bx}\right)\left(\frac{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}}\right) \\
\\ 
 & = \lim_{x \to \infty}\frac{(x^2 + ax) - (x^2 + bx)}{\sqrt{x^2 + ax} + \sqrt{x^2 + bx}} \\ \\
 & = \lim_{x \to \infty}\frac{ax-bx}{\sqrt{x^2\left(1 + \frac{a}{x}\right)} + \sqrt{x^2\left(1 + \frac{b}{x}\right)}} \\ \\
 & = \lim_{x \to \infty}\frac{ax-bx}{|x|\sqrt{1 + \frac{a}{x}} + |x|\sqrt{1 + \frac{b}{x}}} \\ \\
 & = \lim_{x \to \infty}\frac{x(a-b)}{|x|\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}\right)} \\ \\
&\text{Recall:} \ \ \ |x| = x \ \ \ \text{as} \  \ \ x \to \infty \\ \\
 & = \lim_{x \to \infty}\frac{x(a-b)}{x\left(\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}\right)} \\ \\
& = \lim_{x \to \infty}\frac{a-b}{\sqrt{1 + \frac{a}{x}} + \sqrt{1 + \frac{b}{x}}} \\ \\
& = \frac{a-b}{\sqrt{1 + 0} + \sqrt{1 + 0}} \\ \\ 
& = \frac{a-b}{2} \\
\end{split}
\end{equation}
</s>