Ta có \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - 3}}{{2f\left( x \right) + 1}}\)\( = \mathop {\lim }\limits_{x \to a} \frac{{1 - \frac{3}{{f\left( x \right)}}}}{{2 + \frac{1}{{f\left( x \right)}}}} = \frac{{\mathop {\lim }\limits_{x \to a} \left( {1 - \frac{3}{{f\left( x \right)}}} \right)}}{{\mathop {\lim }\limits_{x \to a} \left( {2 + \frac{1}{{f\left( x \right)}}} \right)}}\)

          \( = \frac{{\mathop {\lim }\limits_{x \to a} 1 - \mathop {\lim }\limits_{x \to a} \frac{3}{{f\left( x \right)}}}}{{\mathop {\lim }\limits_{x \to a} 2 + \mathop {\lim }\limits_{x \to a} \frac{1}{{f\left( x \right)}}}} = \frac{{\mathop {\lim }\limits_{x \to a} 1 - \frac{{\mathop {\lim }\limits_{x \to a} 3}}{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}}}{{\mathop {\lim }\limits_{x \to a} 2 + \frac{{\mathop {\lim }\limits_{x \to a} 1}}{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}}}\)\( = \frac{{1 - 0}}{{2 + 0}} = \frac{1}{2}\).

Vậy \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - 3}}{{2f\left( x \right) + 1}} = \frac{1}{2}\).