Highest power of x in denominator is \(\displaystyle{x}^{{2}}\), but since it is beneath square root, we are going to divide both numenator and denominator with x. \(\displaystyle\lim_{{x} \rightarrow \infty}\frac{{{x}−{3}}}{{\sqrt{{4}{x}^{{2}}+{25}}}}:\frac{{x}}{{x}}=\lim_{{x} \rightarrow \infty}\frac{{{1}-{\left(\frac{{3}}{{x}}\right)}}}{{\sqrt{{\left({4}+{\left(\frac{{25}}{{x}^{{2}}}\right)}\right)}}}}=\frac{{{1}-{0}}}{{\sqrt{{4}+{0}}}}=\frac{{1}}{{2}}\)