Đáp án:
$-\dfrac16$
Giải thích các bước giải:
$\quad\lim(\sqrt{n^2 + n-1} - \sqrt[3]{n^3 + 2n^2 +3})$
$=\lim(\sqrt{n^2 + n-1} -n + n- \sqrt[3]{n^3 + 2n^2 +3})$
$=\lim\left[\dfrac{(\sqrt{n^2 + n-1} -n)(\sqrt{n^2 + n -1} + n)}{\sqrt{n^2 + n -1} + n} + \dfrac{( n- \sqrt[3]{n^3 + 2n^2 +3})(n^2 + n\sqrt[3]{n^3 + 2n^2 + 3} + \sqrt[3]{(n^3 + 2n^2 + 3)^2}}{^2 + n\sqrt[3]{n^3 + 2n^2 + 3} + \sqrt[3]{(n^3 + 2n^2 + 3)^2}}\right]$
$=\lim\left[\dfrac{n-1}{\sqrt{n^2 + n+1} + n} - \dfrac{2n^2 +3}{^2 + n\sqrt[3]{n^3 + 2n^2 + 3} + \sqrt[3]{(n^3 + 2n^2 + 3)^2}}\right]$
$=\lim\dfrac{1 - \dfrac1n}{\sqrt{1+ \dfrac1n +\dfrac{1}{n^2}} + 1} - \lim\dfrac{2 +\dfrac{3}{n^2}}{1 + \sqrt[3]{1 + \dfrac2n +\dfrac{3}{n^2}} + \sqrt[3]{\left(1+\dfrac2n +\dfrac{3}{n^2}\right)^2}}$
$=\dfrac{1 - 0}{\sqrt{1 + 0 + 0} + 1} -\dfrac{2 + 3.0}{1 + \sqrt[3]{1 + 2.0 + 3.0} + \sqrt[3]{(1 + 2.0 + 3.0)^2}}$
$=\dfrac12 -\dfrac23$
$=-\dfrac16$
