Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} + \sqrt[3]{{x + 7}} - 4}}{{x - 1}} = \frac{1}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} + \sqrt[3]{{x + 7}} - 4}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {x + 3} - 2} \right) + \left( {\sqrt[3]{{x + 7}} - 2} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\sqrt {x + 3} - 2}}{{x - 1}} + \frac{{\sqrt[3]{{x + 7}} - 2}}{{x - 1}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {\sqrt {x + 3} - 2} \right)\left( {\sqrt {x + 3} + 2} \right)}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}} + \frac{{\left( {\sqrt[3]{{x + 7}} - 2} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + \sqrt[3]{{x + 7}}.2 + {2^2}} \right)}}{{\left( {x - 1} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + \sqrt[3]{{x + 7}}.2 + {2^2}} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{\left( {x + 3} \right) - {2^2}}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}} + \frac{{\left( {x + 7} \right) - {2^3}}}{{\left( {x - 1} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}} + \frac{{x - 1}}{{\left( {x - 1} \right)\left( {{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4} \right)}}} \right]\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{1}{{\sqrt {x + 3} + 2}} + \frac{1}{{{{\sqrt[3]{{x + 7}}}^2} + 2\sqrt[3]{{x + 7}} + 4}}} \right]\\
= \frac{1}{{\sqrt {1 + 3} + 2}} + \frac{1}{{{{\sqrt[3]{{1 + 7}}}^2} + 2\sqrt[3]{{1 + 7}} + 4}}\\
= \frac{1}{3}
\end{array}\)
