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