Đáp án:
4b) $\dfrac{2-\sqrt[3]{6}}{5}$
5c) $4$
Giải thích các bước giải:
4b) $\lim\dfrac{\sqrt[3]{6n^3 +n^2} -2n}{-5n +2}$
$=\lim\dfrac{\sqrt[3]{6 +\dfrac1n} -2}{-5 +\dfrac2n}$
$=\dfrac{\sqrt[3]{6+0} -2}{-5+0}$
$=\dfrac{2-\sqrt[3]{6}}{5}$
5c) $\lim\dfrac{(2n+1)^2}{n^2 - 4n +4}$
$=\lim\dfrac{\left(2+\dfrac1n\right)^2}{1 - \dfrac4n +\dfrac{4}{n^2}}$
$=\dfrac{(2+0)^2}{1-0+0}$
$= 4$
