Tính giới hạn Lim (x^2 +2018)/ (x+1)

\(\mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {{x^2} + 2018} }}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{\left| x \right|\sqrt {1 + \frac{{2018}}{{{x^2}}}} }}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\sqrt {1 + \frac{{2018}}{{{x^2}}}} }}{{x + 1}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {1 + \frac{{2018}}{{{x^2}}}} }}{{1 + \frac{1}{x}}} = 1\).

(Vì \(\mathop {\lim }\limits_{x \to + \infty } \frac{{2018}}{{{x^2}}} = 0;\mathop {\lim }\limits_{x \to + \infty } \frac{1}{x} = 0\)).