Cho \(I = \mathop {\lim }\limits_{x \to 0} \frac{{2(\sqrt {3x + 1}  - 1)}}{x}\) và \(J = \mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} -

* Hướng dẫn giải

\(\begin{array}{l}
I = \mathop {\lim }\limits_{x \to 0} \frac{{2(\sqrt {3x + 1}  - 1)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sqrt {3x + 1}  - 1} \right)\left( {\sqrt {3x + 1}  + 1} \right)}}{{x.\left( {\sqrt {3x + 1}  + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2.3}}{{\sqrt {3x + 1}  + 1}} = 3\\
J = \mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} - x - 2}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \frac{{\left( {x + 1} \right)\left( {x - 2} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to  - 1} \left( {x - 2} \right) =  - 3\\
 \Rightarrow I - J = 6
\end{array}\)