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