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