Đáp án:
21. $1$
22. $\dfrac12$
Giải thích các bước giải:
$\begin{array}{l}21.\quad \lim\left[\dfrac{1}{1.2} + \dfrac{1}{2.3} + \cdots + \dfrac{1}{n(n+1)} \right]\\ = \lim\left(1-\dfrac{1}{2} + \dfrac{1}{2}-\dfrac{1}{3} + \cdots + \dfrac{1}{n}-\dfrac{1}{n+1} \right)\\ = \lim\left(1 - \dfrac{1}{n+1}\right)\\ = 1 - \lim\dfrac{1}{n+1}\\ = 1 - \lim\dfrac{\dfrac1n}{1 + \dfrac1n}\\ = 1 - \dfrac{0}{1 +0}\\ = 1\\ 22.\quad \lim\left[\dfrac{1}{1.3} + \dfrac{1}{3.5} + \cdots + \dfrac{1}{(2n-1)(2n+1)} \right]\\ = \lim\left[\dfrac12\left(\dfrac{2}{1.3} + \dfrac{2}{3.5} + \cdots + \dfrac{2}{(2n-1)(2n+1)} \right)\right]\\ \lim\left[\dfrac12\left(1-\dfrac{1}{3} + \dfrac{1}{3} -\dfrac15+ \cdots + \dfrac{1}{2n-1}-\dfrac{1}{2n+1} \right)\right]\\ = \lim\left[\dfrac{1}{2}\left(1 - \dfrac{1}{2n+1}\right)\right]\\ = \dfrac{1}{2}- \lim\dfrac{1}{2(2n+1)}\\ = \dfrac12 - \lim \dfrac{\dfrac1n}{2\left(2 + \dfrac1n\right)}\\ = \dfrac12 -\dfrac{0}{2(2 +0)}\\ = \dfrac12 \end{array}$
