Tính \(\mathop {\lim \to + \infty } {2{x^2} + x} - \sqrt {{x^2} + 1} }}{{2x + 1}}.\)

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}\,\,\,\,\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {2{x^2} + x}  - \sqrt {{x^2} + 1} }}{{2x + 1}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {2 + \frac{1}{x}}  - \sqrt {1 + \frac{1}{x}} }}{{2 + \frac{1}{x}}}\\ = \frac{{\sqrt 2  - 1}}{2}\end{array}\)