Đáp án:
a) $\lim\dfrac{\sin n}{n} =0$
b) $\lim\left[\dfrac{(-1)^n}{n+1} + 1\right] =1$
Giải thích các bước giải:
$\begin{array}{l}1)\quad \lim\dfrac{\sin n}{n}\\ \text{Ta có:}\\ \quad -\dfrac1n \leq \dfrac{\sin n}{n} \leq \dfrac{1}{n}\\ \to \lim\left(-\dfrac1n\right) \leq \lim\dfrac{\sin n}{n} \leq \lim\dfrac1n\\ mà\,\,\lim\left(-\dfrac1n\right) = \lim\dfrac1n = 0\\ nên\,\,\lim\dfrac{\sin }{n} = 0\quad \text{(Định lý giới hạn kẹp)}\\ 2)\quad \lim\left[\dfrac{(-1)^n}{n+1} + 1\right]\\ = \lim\dfrac{(-1)^n}{n+1} + 1\\ \text{Ta có:}\ -\dfrac1n < -\dfrac{1}{n+1} \leq \dfrac{(-1)^n}{n+1} \leq \dfrac{1}{n+1} < \dfrac1n\\ \to \lim\left(-\dfrac1n\right) < \lim\dfrac{(-1)^n}{n+1} < \lim\dfrac1n\\ mà\,\,\lim\left(-\dfrac1n\right) = \lim\dfrac1n = 0\\ nên\,\,\lim\dfrac{(-1)^n}{n+1} =0\quad \text{(Định lý giới hạn kẹp)}\\ Vậy\,\,\lim\left[\dfrac{(-1)^n}{n+1} + 1\right] = 0 + 1 = 1 \end{array}$
