\[\begin{array}{l}
\mathop {\lim }\limits_{x \to 9} \dfrac{{\sqrt {x + 7} - 4}}{{\sqrt x - 3}} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\left( {\sqrt {x + 7} - 4} \right)\left( {\sqrt {x + 7} + 4} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)\left( {\sqrt {x + 7} + 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 9} \dfrac{{\left( {x - 9} \right)\left( {\sqrt x + 3} \right)}}{{\left( {x - 9} \right)\left( {\sqrt {x + 7} + 4} \right)}} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\sqrt x + 3}}{{\sqrt {x + 7} + 4}}\\
= \dfrac{{\sqrt 9 + 3}}{{\sqrt {16} + 4}} = \dfrac{6}{8} = \dfrac{3}{4}\\
\mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt {1 + 2x} - 3}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \dfrac{{\left( {\sqrt {1 + 2x} - 3} \right)\left( {\sqrt {1 + 2x} + 3} \right)}}{{\left( {x - 4} \right)\left( {\sqrt {1 + 2x} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{2x - 8}}{{\left( {x - 4} \right)\left( {\sqrt {1 + 2x} + 3} \right)}} = \mathop {\lim }\limits_{x \to 4} \dfrac{2}{{\left( {\sqrt {1 + 2x} + 3} \right)}} = \dfrac{2}{{3 + 3}} = \dfrac{1}{3}
\end{array}\]
