Đáp án:
\[\mathop {\lim }\limits_{x \to - 4} \frac{{x + \sqrt {x + 20} }}{{\sqrt {2x + 9} - 1}} = \frac{9}{8}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to - 4} \frac{{x + \sqrt {x + 20} }}{{\sqrt {2x + 9} - 1}}\\
= \mathop {\lim }\limits_{x \to - 4} \frac{{\left( {x + 4} \right) + \left( {\sqrt {x + 20} - 4} \right)}}{{\sqrt {2x + 9} - 1}}\\
= \mathop {\lim }\limits_{x \to - 4} \dfrac{{x + 4 + \frac{{\left( {x + 20} \right) - {4^2}}}{{\sqrt {x + 20} + 4}}}}{{\frac{{2x + 9 - {1^2}}}{{\sqrt {2x + 9} + 1}}}}\\
= \mathop {\lim }\limits_{x \to - 4} \dfrac{{x + 4 + \frac{{x + 4}}{{\sqrt {x + 20} + 4}}}}{{\frac{{2\left( {x + 4} \right)}}{{\sqrt {2x + 9} + 1}}}}\\
= \mathop {\lim }\limits_{x \to - 4} \frac{{\left( {x + 4} \right)\left( {1 + \frac{1}{{\sqrt {x + 20} + 4}}} \right)}}{{\left( {x + 4} \right).\frac{2}{{\sqrt {2x + 9} + 1}}}}\\
= \mathop {\lim }\limits_{x \to - 4} \dfrac{{1 + \frac{1}{{\sqrt {x + 20} + 4}}}}{{\frac{2}{{\sqrt {2x + 9} + 1}}}}\\
= \dfrac{{1 + \frac{1}{{\sqrt { - 4 + 20} + 4}}}}{{\frac{2}{{\sqrt {2.\left( { - 4} \right) + 9} + 1}}}}\\
= \dfrac{{1 + \frac{1}{8}}}{{\frac{2}{{\sqrt 1 + 1}}}}\\
= \frac{9}{8}
\end{array}\)
