Kết quả lim x → + ∞ ( √ 2020 x 2 + x + 3 − √ 2021 x 2 + 2 ) bằng

Ta có\[\mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left( {\sqrt {2020{{\rm{x}}^2} + {\rm{x}} + 3} - \sqrt {2021{{\rm{x}}^2} + 2} } \right) = \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left[ {{\rm{x}}\left( {\sqrt {2020 + \frac{1}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right)} \right]\]

Vì\[\mathop {\lim }\limits_{{\rm{x}} \to + \infty } {\rm{x}} = + \infty ,\mathop {\lim }\limits_{{\rm{x}} \to \infty } \left( {\sqrt {2020 + \frac{1}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right) = \sqrt {2020} - \sqrt {2021} < 0\]

\[ \Rightarrow \mathop {\lim }\limits_{{\rm{x}} \to + \infty } \left[ {{\rm{x}}\left( {\sqrt {2020 + \frac{{\rm{1}}}{{\rm{x}}} + \frac{3}{{{{\rm{x}}^2}}}} - \sqrt {2021 + \frac{2}{{{{\rm{x}}^2}}}} } \right)} \right] = - \infty \]

Chọn đáp án A