as x approaches positive infinity
the square root (in parenthesis) of (x ^2 + ax) minus x
Thanks in advance.
Have you learned L'Hospital's rule? We can occasionally rewrite a limit of the form $\displaystyle \infty - \infty$ as a limit of the form $\displaystyle \frac{\infty}{\infty}$. In our case,
$\displaystyle \begin{aligned}
\lim_{\small x\rightarrow\infty}(\sqrt{x^2+ax}-x)&=\lim_{\small x\rightarrow\infty}\left(x\sqrt{1+\frac{a}{x}}-x\right)\\
&=\lim_{\small x\rightarrow\infty}\left(x\left(\sqrt{1+\frac{a}{x }}-1\right)\right)\\
&=\lim_{\small x\rightarrow\infty}\frac{\sqrt{1+\frac{a}{x}}-1}{\frac{1}{x}}.
\end{aligned}
$
Edit: Or you can use Mr. Fantastic's method, which doesn't require the rule.