Đáp án:
$\lim(2n-\sqrt{9n^2 +n} +\sqrt{n^2 + 2n})=\dfrac56$
Giải thích các bước giải:
$\quad \lim(2n-\sqrt{9n^2 +n} +\sqrt{n^2 + 2n})$
$=\lim(3n-\sqrt{9n^2 +n} +\sqrt{n^2 + 2n}-n)$
$=\lim(3n-\sqrt{9n^2 +n}) + \lim(\sqrt{n^2 + 2n}-n)$
$=\lim\dfrac{(3n-\sqrt{9n^2 +n})(3n+\sqrt{9n^2 +n})}{3n+\sqrt{9n^2 +n}} + \lim\dfrac{(\sqrt{n^2 + 2n}-n)(\sqrt{n^2 + 2n}+n)}{\sqrt{n^2 + 2n}+n}$
$=\lim\dfrac{9n^2 - (9n^2 +n)}{3n+\sqrt{9n^2 +n}} + \lim\dfrac{n^2 + 2n -n^2}{\sqrt{n^2 + 2n}+n}$
$=\lim\dfrac{-n}{3n+\sqrt{9n^2 +n}} + \lim\dfrac{2n}{\sqrt{n^2 + 2n}+n}$
$=\lim\dfrac{-1}{3+\sqrt{9 +\dfrac1n}} + \lim\dfrac{2}{\sqrt{1 + \dfrac2n}+1}$
$=\dfrac{-1}{3+\sqrt{9+0}}+\dfrac{2}{\sqrt{1+0} +1}$
$= -\dfrac16 + 1$
$=\dfrac56$