Đáp án:
$\lim\left(\sqrt{n^2 + 4n -3} - n - 5\right)=-3$
Giải thích các bước giải:
$\quad \lim\left(\sqrt{n^2 + 4n -3} - n - 5\right)$
$=\lim\dfrac{\left(\sqrt{n^2 + 4n -3} - n - 5\right)\left(\sqrt{n^2 + 4n -3} + n + 5\right)}{\sqrt{n^2 + 4n -3} + n +5}$
$=\lim\dfrac{n^2 + 4n +3 - (n+5)^2}{\sqrt{n^2 + 4n -3} + n +5}$
$=\lim\dfrac{-6n-22}{\sqrt{n^2 + 4n -3} + n +5}$
$=\lim\dfrac{-6-\dfrac{22}{n}}{\sqrt{1+\dfrac4n -\dfrac{3}{n^2}}+1}$
$=\dfrac{-6-0}{\sqrt{1+0-0} +1}$
$= -3$
