Đáp án:
Giải thích các bước giải:
i) $\lim\dfrac{\sqrt{4n^2 +1} - 2n -1}{\sqrt{n^2 + 4n +1} - n}$
$=\lim\dfrac{(\sqrt{4n^2 +1} - 2n -1)(\sqrt{4n^2 +1} +2n +1)(\sqrt{n^2 + 4n +1} + n)}{(\sqrt{4n^2 +1} +2n +1)(\sqrt{n^2 + 4n +1} +n)(\sqrt{n^2 + 4n +1} - n)}$
$=\lim\dfrac{- 4n(\sqrt{n^2 + 4n +1} + n)}{(4n+1)(\sqrt{4n^2 +1} + 2n+1)}$
$=-\lim\dfrac{4\left(\sqrt{1 + \dfrac4n +\dfrac{1}{n^2}} + 1\right)}{\left(4 +\dfrac1n\right)\left(\sqrt{4 +\dfrac{1}{n^2}} + 2 +\dfrac1n\right)}$
$= -\dfrac{4(\sqrt{1 +0 + 0} + 1)}{(4 +0)(\sqrt{4 + 0} + 2 + 0)}$
$= -\dfrac12$
q) $\lim\left[\dfrac{1}{1.2} +\dfrac{1}{2.3} +\cdots +\dfrac{1}{n(n+1)}\right]$
$=\lim\left(1 -\dfrac12 +\dfrac12 -\dfrac13 +\cdots + \dfrac1n -\dfrac{1}{n+1}\right)$
$=\lim\left(1 -\dfrac{1}{n+1}\right)$
$= 1 -\lim\dfrac{1}{n+1}$
$= 1 - 0$
$= 1$
