Đáp án:
\(\dfrac{8}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } \dfrac{{9{x^2} + 16x + 2 - 9\left( {1 + {x^2}} \right)}}{{\sqrt {9{x^2} + 16x + 2} + 3\sqrt {1 + {x^2}} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{16x - 7}}{{\sqrt {9{x^2} + 16x + 2} + 3\sqrt {1 + {x^2}} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{16 - \dfrac{7}{x}}}{{\sqrt {9 + \dfrac{{16}}{x} + \dfrac{2}{{{x^2}}}} + 3\sqrt {\dfrac{1}{{{x^2}}} + 1} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{{16}}{{3 + 3}} = \dfrac{8}{3}
\end{array}\)
