Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to 2} \dfrac{{3x + 5}}{{x + 1}} = \dfrac{{3.2 + 5}}{{2 + 1}} = \dfrac{{11}}{3}\\
b,\\
\mathop {\lim }\limits_{x \to 0} \dfrac{{2 - \sqrt {4 - x} }}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {2 - \sqrt {4 - x} } \right)\left( {2 + \sqrt {4 - x} } \right)}}{{x.\left( {2 + \sqrt {4 - x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{2^2} - \left( {4 - x} \right)}}{{x.\left( {2 + \sqrt {4 - x} } \right)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{x.\left( {2 + \sqrt {4 - x} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{1}{{2 + \sqrt {4 - x} }} = \dfrac{1}{{2 + \sqrt {4 - 0} }} = \dfrac{1}{4}
\end{array}\)
