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

* Hướng dẫn giải

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