Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {4x + 5} - 3}}{{\sqrt[3]{{5x + 3}} - 2}} = \frac{8}{5}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {4x + 5} - 3}}{{\sqrt[3]{{5x + 3}} - 2}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {4x + 5} - 3} \right)\left( {\sqrt {4x + 5} + 3} \right)\left( {{{\sqrt[3]{{5x + 3}}}^2} + 2.\sqrt[3]{{5x + 3}} + {2^2}} \right)}}{{\left( {\sqrt[3]{{5x + 3}} - 2} \right)\left( {{{\sqrt[3]{{5x + 3}}}^2} + 2.\sqrt[3]{{5x + 3}} + {2^2}} \right)\left( {\sqrt {4x + 5} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left[ {{{\sqrt {4x + 5} }^2} - {3^2}} \right].\left( {{{\sqrt[3]{{5x + 3}}}^2} + 2\sqrt[3]{{5x + 3}} + 4} \right)}}{{\left( {{{\sqrt[3]{{5x + 3}}}^3} - {2^3}} \right)\left( {\sqrt {4x + 5} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {4x + 5 - 9} \right)\left( {{{\sqrt[3]{{5x + 3}}}^2} + 2.\sqrt[3]{{5x + 3}} + 4} \right)}}{{\left( {5x + 3 - 8} \right)\left( {\sqrt {4x + 5} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{4\left( {x - 1} \right)\left( {{{\sqrt[3]{{5x + 3}}}^2} + 2\sqrt[3]{{5x + 3}} + 4} \right)}}{{5\left( {x - 1} \right)\left( {\sqrt {4x + 5} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{4\left( {{{\sqrt[3]{{5x + 3}}}^2} + 2\sqrt[3]{{5x + 3}} + 4} \right)}}{{5\left( {\sqrt {4x + 5} + 3} \right)}}\\
= \frac{{4.\left( {{{\sqrt[3]{{5.1 + 3}}}^2} + 2.\sqrt[3]{{5.1 + 3}} + 4} \right)}}{{5.\left( {\sqrt {4.1 + 5} + 3} \right)}}\\
= \frac{{4.12}}{{5.6}} = \frac{8}{5}
\end{array}\)
Vậy \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {4x + 5} - 3}}{{\sqrt[3]{{5x + 3}} - 2}} = \frac{8}{5}\)
