Use law 8 and law 3 the properties of limit.
\(\begin{array}{c}\frac{{\mathop {\lim }\limits_{x \to \infty } 4 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } 5 - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right)}} = \frac{{\mathop {\lim }\limits_{x \to \infty } 4 + \mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } 5 - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right)}}\\ = \frac{{4 + 3\left( 0 \right)}}{{5 - 1\left( 0 \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{Using}}\,{\rm{theorem 5}}} \right)\\ = \frac{4}{5}\end{array}\)
So, the value of the limit is \(\frac{4}{5}\).