Answer
$\dfrac{1}{2}$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{y \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of a limit, so we need to use L'Hospital's rule.
Here, $A'(x)=\dfrac{1}{2}(ay+a^2)^{-1/2}(a)=\dfrac{a}{2(ay+a^2)^{1/2}}$ and $B'(x)=1$
$ \lim\limits_{y \to 0} \dfrac{\dfrac{a}{2(ay+a^2)^{1/2}}}{1}=\dfrac{a}{2\sqrt {a(0)+a^2}}=\dfrac{1}{2}$
