Tính \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^2} - x + 3} }}{{2\left| x \right| - 1}}\) bằng:

* Hướng dẫn giải

\(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^2} - x + 3} }}{{2\left| x \right| - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{x\left( {2 - \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{2 - \frac{1}{x}}} = 3\)