Lời giải:Ta có \(\int\limits_1^k {\left( {2x - 1} \right)dx} = \left. {\left( {{x^2} - x} \right)} \right|_1^k = {k^2} - k\)
Mà \(4\mathop {\lim }\limits_{x \to 0} = \frac{{\sqrt {x + 1} - 1}}{x} = 4\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}}{{x\left( {\sqrt {x + 1} + 1} \right)}} = 4\mathop {\lim }\limits_{x \to 0} \frac{1}{{\sqrt {x + 1} + 1}} = 2\)
Khi đó \(\int\limits_1^k {\left( {2x - 1} \right)dx} = 4\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - 1}}{x} \Leftrightarrow {k^2} - k = 2 \Leftrightarrow \left[ \begin{array}{l}k = 2\\k = - 1\end{array} \right.\)
