Đáp án:
g) $\lim\dfrac{1+2+\dots n}{n^2}=\dfrac12$
h) $\lim(\sqrt{n^2+1} -\sqrt[3]{n^3+n^2})=-\dfrac13$
Giải thích các bước giải:
g) $\lim\dfrac{1+2+\dots n}{n^2}$
$=\lim\dfrac{n(n+1)}{2n^w}$
$=\lim\dfrac{n+1}{2n}$
$=\lim\dfrac{1+\dfrac1n}{2}$
$=\dfrac{1+0}{2}$
$=\dfrac12$
h) $\lim(\sqrt{n^2+1} -\sqrt[3]{n^3+n^2})$
$=\lim(\sqrt{n^2+1} -n + n-\sqrt[3]{n^3+n^2})$
$=\lim(\sqrt{n^2+1} -n) +\lim(n-\sqrt[3]{n^3+n^2})$
$=\lim\dfrac{(\sqrt{n^2+1} -n)(\sqrt{n^2+1} +n)}{\sqrt{n^2+1} +n} +\lim\dfrac{(n-\sqrt[3]{n^3+n^2})(n^2 + n\sqrt[3]{n^3 + n^2} +\sqrt[3]{(n^3 +n^2)^2})}{n^2 + n\sqrt[3]{n^3 + n^2} +\sqrt[3]{(n^3 +n^2)^2}}$
$=\lim\dfrac{1}{\sqrt{n^2+1} +n} +\lim\dfrac{-n^2}{n^2 + n\sqrt[3]{n^3 + n^2} +\sqrt[3]{(n^3 +n^2)^2}}$
$=\lim\dfrac{\dfrac1n}{\sqrt{1+\dfrac1n} +1} +\lim\dfrac{-1}{1 + \sqrt[3]{1 +\dfrac1n} +\sqrt[3]{\left(1+\dfrac1n\right)^2}}$
$=\dfrac{0}{\sqrt{1+0} +1} -\dfrac{1}{1 + \sqrt[3]{1 +0} +\sqrt[3]{\left(1+0\right)^2}}$
$=0-\dfrac{1}{1+1+1}$
$=-\dfrac13$
