\(\lim\dfrac{2-4n}{3n+1}=\text{lim}\dfrac{n\left(\dfrac{2}{n}-4\right)}{n\left(3+\dfrac{1}{n}\right)}=\lim\dfrac{\dfrac{2}{n}-4}{3+\dfrac{1}{n}}=\dfrac{0-4}{3+0}=-\dfrac{4}{3}\)
\(\lim\dfrac{2n^4+3n^2-2}{1-n^2}=\lim\dfrac{n^2\left(2n^2+3-\dfrac{2}{n^2}\right)}{n^2\left(\dfrac{1}{n^2}-1\right)}=\lim\dfrac{2n^2+3-\dfrac{2}{n^2}}{\dfrac{1}{n^2}-1}=\dfrac{+\infty+3-0}{0-1}=-\infty\)
Đúng 1
Bình luận (0)