Thomas' Calculus 13th Edition

$\dfrac{1}{2}$
Here, we have $\lim\limits_{y \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ Now, $p'(x)=(\dfrac{1}{2})(ay+a^2)^{-1/2}(a)=\dfrac{a}{2(ay+a^2)^{1/2}}$ and $q'(x)=1$ $\lim\limits_{y \to 0} \dfrac{\dfrac{a}{2(ay+a^2)^{1/2}}}{1}=\dfrac{a}{2\sqrt {a(0)+a^2}}$ or, $\dfrac{a}{2a}=\dfrac{1}{2}$