Tìm lim (2020.n^2−n)/(2021+n^2)

Chia tử số và mẫu số cho n², ta được $\frac{2020{\mathrm{n}}^2-\mathrm{n}}{2021+{\mathrm{n}}^2}=\frac{2020-\frac{1}{\mathrm{n}}}{\frac{2021}{{\mathrm{n}}^2}+1}$

Vì $\lim \left(2020-\frac{1}{\mathrm{n}}\right)=\lim 2020-\lim \frac{1}{\mathrm{n}}$$=2020-0=2020$

Và $\lim \left(2021.\frac{1}{\mathrm{n}}\frac{1}{\mathrm{n}}+1\right)$$=\lim 2021.\lim \frac{1}{\mathrm{n}}.\lim \frac{1}{\mathrm{n}}+\lim 1$$=2021.0.0+1=1$

Nên $\lim \frac{2020{\mathrm{n}}^2-\mathrm{n}}{2021+{\mathrm{n}}^2}=\lim \frac{2020-\frac{1}{\mathrm{n}}}{\frac{2021}{{\mathrm{n}}^2}+1}$$=\frac{\lim \left(2020-\frac{1}{\mathrm{n}}\right)}{\lim \left(\frac{2021}{{\mathrm{n}}^2}+1\right)}$$=\frac{2020}{1}$$=2020$

Vậy $\lim \frac{2020{\mathrm{n}}^2-\mathrm{n}}{2021+{\mathrm{n}}^2}=2020$

Chọn đáp án D.