Ai giúp mình bài này vs

Đáp án:

$2)\\ a) \dfrac{5}{3}\\ b) -\dfrac{3}{2}\\ c)0\\ d)+\infty\\ e)-\infty\\ f) +\infty.$

Giải thích các bước giải:

$2)\\ a) \lim \dfrac{5n-1}{3n+2}\\ =\lim \dfrac{5-\dfrac{1}{n}}{3+\dfrac{2}{n}}\\ =\dfrac{5}{3}\\ b) \lim \dfrac{3n^2+n-1}{4-2n^2}\\ = \lim \dfrac{3+\dfrac{1}{n}-\dfrac{1}{n^2}}{\dfrac{4}{n^2}-2}\\ =-\dfrac{3}{2}\\ c)\lim \dfrac{n+1}{n^2+n-2}\\ =\lim \dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1+\dfrac{1}{n}-\dfrac{2}{n^2}}\\ =0\\ d)\lim (4n^3+2n-1)\\ =\lim n^3\left(4+\dfrac{2}{n^2}-\dfrac{1}{n^3}\right)\\ =\lim n^3.\lim\left(4+\dfrac{2}{n^2}-\dfrac{1}{n^3}\right)\\ =+\infty\\ e) \lim (4+n-n^2)\\ = \lim n^2 \left(\dfrac{4}{n^2}+\dfrac{1}{n}-1\right)\\ =\lim n^2.\lim \left(\dfrac{4}{n^2}+\dfrac{1}{n}-1\right)\\ =-\infty\\ f) \lim \left(\sqrt{4n^2-n+1}-n\right)\\\ =\lim n\left(\sqrt{4-\dfrac{1}{n}+\dfrac{1}{n^2}}-1\right)\\ =\lim n.\lim\left(\sqrt{4-\dfrac{1}{n}+\dfrac{1}{n^2}}-1\right)\\ =+\infty.$