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

* Hướng dẫn giải

\(\begin{array}{l}
I = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1}  - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\sqrt {2x + 1}  + 1}} = 1\\
J = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + x - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)(x + 2)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} (x + 2) = 3\\
 \Rightarrow I + J = 4
\end{array}\)