Answer
$$\frac{1}{2}$$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow \infty} \frac{x-3}{\sqrt{4 x^{2}+25}}&=\lim _{x \rightarrow \infty} \frac{(x-3) / \sqrt{x^{2}}}{\sqrt{4 x^{2}+25} / \sqrt{x^{2}}}\\
&=\lim _{x \rightarrow \infty} \frac{(x-3) / x}{\sqrt{\left(4 x^{2}+25\right) / x^{2}}}\\
&=\lim _{x \rightarrow \infty} \frac{(1-3 / x)}{\sqrt{4+25 / x^{2}}}\\
&=\frac{(1-0)}{\sqrt{4+0}}\\
&=\frac{1}{2}
\end{align*}
