Đáp án:
a) $-\infty$
b) $0$
c) $-\infty$
Giải thích các bước giải:
a) $\lim\dfrac{-n^5 + 3n +2}{3n+5}$
$=\lim\left[n^4\left(\dfrac{-1 +\dfrac{3}{n^3} + \dfrac{2}{n^4}}{3 +\dfrac5n}\right)\right]$
$=+\infty \cdot \dfrac{-1 +0 + 0}{3 + 0}$
$= -\infty$
b) $\lim\dfrac{n^2 - 2n+1}{n^6 - 3n +2}$
$=\lim\dfrac{\dfrac{1}{n^4} -\dfrac{2}{n^5} +\dfrac{1}{n^6}}{1 -\dfrac{3}{n^5} +\dfrac{2}{n^6}}$
$=\dfrac{0 - 0+ 0}{1 - 0 + 0}$
$= 0$
c) $\lim(n^3 - 2n + n^6 - n^7)$
$=\lim\left[n^7\left(\dfrac{1}{n^4} - \dfrac{2}{n^6} +\dfrac1n -1\right)\right]$
$=+\infty (0 - 0 +0 -1)$
$= -\infty$
