Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
\mathop {\lim }\limits_{x \to 2} \dfrac{{2 - x}}{{\sqrt {x + 7} - 3}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2 - x} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{\left( {\sqrt {x + 7} - 3} \right)\left( {\sqrt {x + 7} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2 - x} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{\left( {x + 7} \right) - {3^2}}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {2 - x} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{x - 2}}\\
= \mathop {\lim }\limits_{x \to 2} \left[ { - \left( {\sqrt {x + 7} + 3} \right)} \right] = - \left( {\sqrt {2 + 7} + 3} \right) = - 6\\
6,\\
\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {4x + 1} - 3}}{{{x^2} - 4}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {\sqrt {4x + 1} - 3} \right)\left( {\sqrt {4x + 1} + 3} \right)}}{{\left( {{x^2} - 4} \right)\left( {\sqrt {4x + 1} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {4x + 1} \right) - {3^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{4\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{4}{{\left( {x + 2} \right)\left( {\sqrt {4x + 1} + 3} \right)}}\\
= \dfrac{4}{{\left( {2 + 2} \right).\left( {\sqrt {4.2 + 1} + 3} \right)}} = \dfrac{1}{6}\\
7,\\
\mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt {x + 5} - \sqrt {2x + 1} }}{{x - 4}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{\left( {\sqrt {x + 5} - \sqrt {2x + 1} } \right)\left( {\sqrt {x + 5} + \sqrt {2x + 1} } \right)}}{{\left( {x - 4} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{\left( {x + 5} \right) - \left( {2x + 1} \right)}}{{\left( {x - 4} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{4 - x}}{{\left( {x - 4} \right)\left( {\sqrt {x + 5} + \sqrt {2x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{ - 1}}{{\sqrt {x + 5} + \sqrt {2x + 1} }}\\
= \dfrac{{ - 1}}{{\sqrt {4 + 5} + \sqrt {2.5 + 1} }} = - \dfrac{1}{6}\\
8,\\
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 1} + \sqrt {x + 4} - 3}}{x}\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sqrt {x + 1} - 1}}{x} + \dfrac{{\sqrt {x + 4} - 2}}{x}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}}{{x\left( {\sqrt {x + 1} + 1} \right)}} + \dfrac{{\left( {\sqrt {x + 4} - 2} \right)\left( {\sqrt {x + 4} + 2} \right)}}{{x\left( {\sqrt {x + 4} + 2} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\left( {x + 1} \right) - {1^2}}}{{x\left( {\sqrt {x + 1} + 1} \right)}} + \dfrac{{\left( {x + 4} \right) - {2^2}}}{{x\left( {\sqrt {x + 4} + 2} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{x}{{x\left( {\sqrt {x + 1} + 1} \right)}} + \dfrac{x}{{x\left( {\sqrt {x + 4} + 2} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{1}{{\sqrt {x + 1} + 1}} + \dfrac{1}{{\sqrt {x + 4} + 2}}} \right]\\
= \dfrac{1}{{\sqrt {0 + 1} + 1}} + \dfrac{1}{{\sqrt {0 + 4} + 2}}\\
= \dfrac{3}{4}
\end{array}\)
