Biết lim (2n^3+n^2−4)/(2+n+4n^3) = L. Khi đó 1 – L^2 bằng

* Hướng dẫn giải

$\lim \frac{2{\mathrm{n}}^3+{\mathrm{n}}^2-4}{2+\mathrm{n}+4{\mathrm{n}}^3}=\lim \frac{{\mathrm{n}}^3\left(2+\frac{1}{\mathrm{n}}-\frac{4}{{\mathrm{n}}^3}\right)}{{\mathrm{n}}^3\left(\frac{2}{{\mathrm{n}}^3}+\frac{1}{{\mathrm{n}}^2}+4\right)}$$=\frac{2}{4}=\frac{1}{2}$

Suy ra $\mathrm{L}=\frac{1}{2}$. Khi đó $1-{\mathrm{L}}^2=1-{\left(\frac{1}{2}\right)}^2=\frac{3}{4}$