Đáp án:
$\lim\dfrac{1}{\sqrt{2n+1} -\sqrt{n+1}}=0$
Giải thích các bước giải:
$\lim\dfrac{1}{\sqrt{2n+1} -\sqrt{n+1}}$
$=\lim\dfrac{\sqrt{2n+1} +\sqrt{n+1}}{(\sqrt{2n+1} -\sqrt{n+1})(\sqrt{2n+1} +\sqrt{n+1})}$
$=\lim\dfrac{\sqrt{2n+1} +\sqrt{n+1}}{n}$
$=\lim\left(\sqrt{\dfrac2n +\dfrac{1}{n^2}} +\sqrt{\dfrac1n +\dfrac{1}{n^2}}\right)$
$=\sqrt{2.0 + 0} + \sqrt{0+0}$
$=0$
