Giải

a) $\displaystyle\lim\dfrac{n^2 - 2n + 1}{2 - 3n^2} = \lim\dfrac{1 - \frac{2}{n} + \frac{1}{n^2}}{\frac{2}{n^2} - 3} = \lim\dfrac{1 - 0 + 0}{0 - 3} = -\dfrac{1}{3}.$

b) $\displaystyle\lim\dfrac{2n^2 + n - 3}{n^3 + 5} = \lim\dfrac{2 + \frac{1}{n} - \frac{3}{n^2}}{n + \frac{5}{n^2}} = \lim\dfrac{0 + 0 - 0}{1 + 0} = 0.$

c) $\displaystyle\lim(\sqrt{n^2 + 2n} - n) = \lim\dfrac{(\sqrt{n^2 + 2n} - n)(\sqrt{n^2 + 2n} + n)}{\sqrt{n^2 + 2n} + n} = \lim\dfrac{2n}{\sqrt{n^2 + 2n} + n} = \lim\dfrac{2}{\sqrt{1 + \frac{2}{n}} + 1} = 1.$

Giải

a) $\displaystyle\lim\dfrac{3^{n+1}}{2^{2n}} = \lim\dfrac{3\cdot3^n}{4^n} = 3\cdot\lim\left(\dfrac{3}{4}\right)^n = 3\cdot0 = 0$ (vì $\frac{3}{4} < 1$).

b) $\displaystyle\lim\dfrac{3^{n+1} + 2^n}{3^n} = \lim(3 + \left(\dfrac{2}{3}\right)^n) = 3 + 0 = 3.$

Giải

a) $n^2 + 3n - 5 = n^2\left(1 + \dfrac{3}{n} - \dfrac{5}{n^2}\right)$ $\Rightarrow \lim n^2 = +\infty$ và $\lim\left(1 + \dfrac{3}{n} - \dfrac{5}{n^2}\right) = 1 + 0 - 0 = 1.$ $\Rightarrow \lim(n^2 + 3n - 5) = +\infty.$

b) $\displaystyle\dfrac{n^2 + 7}{1 - 2n} = \dfrac{n^2\left(1 + \frac{7}{n^2}\right)}{n\left(\frac{1}{n} - 2\right)} = \dfrac{n(1 + \frac{7}{n^2})}{\frac{1}{n} - 2}.$ Ta có $\lim n = +\infty$ và $\lim\dfrac{1 + \frac{7}{n^2}}{\frac{1}{n} - 2} = \dfrac{1 + 0}{0 - 2} = -\dfrac{1}{2}.$ $\Rightarrow \lim\dfrac{n^2 + 7}{1 - 2n} = -\infty.$

c) $3^n - 2^n = 3^n\left[1 - \left(\dfrac{2}{3}\right)^n\right]$ $\Rightarrow \lim 3^n = +\infty$, $\lim\left[1 - \left(\dfrac{2}{3}\right)^n\right] = 1 - 0 = 1.$ $\Rightarrow \lim(3^n - 2^n) = +\infty.$