In agreement with all the comments that appear so quickly,
$$ \begin{align} \tag1\lim_{b\to+\infty}\left(\sqrt{b^2+ab}-b\right)&=\lim_{b\to+\infty}\frac{\left(\sqrt{b^2+ab}-b\right)\left(\sqrt{b^2+ab}+b\right)}{\sqrt{b^2+ab}+b}\\ \tag2&=\lim_{b\to+\infty}\frac{b^2+ab-b^2}{\sqrt{b^2+ab}+b}\\ \tag3&=\lim_{b\to+\infty}\frac{a}{\sqrt{1+\frac ab}+1}\\ \tag4&=\frac a2 \end {align} $$ where there is no step that is controversial.
There was briefly a deleted answer that had a mistake in it. Upon second thought, there must be a way to fix it, and so here goes: Let $y=\sqrt{b^2+ab}-b$
$$ \begin{align} \tag5y+b&=\sqrt{b^2+ab}\\ \tag6y^2+2yb+b^2&=b^2+ab\\ \tag7y^2+2yb&=ab\\ \tag8\frac{y^2}b+2y&=a \end {align} $$ This is an equality that always holds, and so if we can show that $\lim_{b\to+\infty}\frac{y^2}b=0$ then the result $y\to\frac a2$ will become self-evident.
Thus, we soldier on, $$ \begin{align} \tag9y^2&=b^2+ab+b^2-2b\sqrt{b^2+ab}\\ \tag{10}\frac{y^2}b&=2b+a-2\sqrt{b^2+ab}\\ \tag{11}&=\frac{(2b+a)^2-4(b^2+ab)}{2b+a+2\sqrt{b^2+ab}}\\ \tag{12}&=\frac{a^2}{2b+a+2\sqrt{b^2+ab}}\\ \tag{13}\therefore\qquad\lim_{b\to+\infty}\frac{y^2}b&=0 \end {align} $$ But of course, this is just a longer route to the same result.
