## Thomas' Calculus 13th Edition

$\dfrac{1}{2}$
Here, $\lim\limits_{y \to 0} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ we have $p'(x)=\dfrac{1}{2}(5y+25)^{-1/2}(5)=\dfrac{5}{2(5+25)^{1/2}}$ and $q'(x)=1$ $\lim\limits_{y \to 0} \dfrac{\dfrac{5}{2(5+25)^{1/2}}}{1}=\dfrac{5}{2\sqrt {5(0)+25}}$ or, $\dfrac{5}{10}=\dfrac{1}{2}$