Đáp án:
\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} + \sqrt {{x^2} + 3x} - 4}}{{x - 1}} = \frac{3}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} + \sqrt {{x^2} + 3x} - 4}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {x + 3} - 2} \right) + \left( {\sqrt {{x^2} + 3x} - 2} \right)}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{\left( {x + 3} \right) - {2^2}}}{{\sqrt {x + 3} + 2}} + \frac{{\left( {{x^2} + 3x} \right) - {2^2}}}{{\sqrt {{x^2} + 3x} + 2}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{x - 1}}{{\sqrt {x + 3} + 2}} + \frac{{{x^2} + 3x - 4}}{{\sqrt {{x^2} + 3x} + 2}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{x - 1}}{{\sqrt {x + 3} + 2}} + \frac{{\left( {x - 1} \right)\left( {x + 4} \right)}}{{\sqrt {{x^2} + 3x} + 2}}}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{1}{{\sqrt {x + 3} + 2}} + \frac{{x + 4}}{{\sqrt {{x^2} + 3x} + 2}}} \right]\\
= \frac{1}{{\sqrt {1 + 3} + 2}} + \frac{{1 + 4}}{{\sqrt {{1^2} + 3.1} + 2}}\\
= \frac{1}{4} + \frac{5}{4}\\
= \frac{3}{2}
\end{array}\)
