Tính các giới hạn sau: $1)$ $lim(n-\sqrt{n^2+3})$ $2)$ $lim(\sqrt[3]{n^2-n^3}+n)$

Đáp án:$1)\lim(n-\sqrt{n^2+3})=0$

$2)\lim(\sqrt[3]{n^2-n^3}+n)=\dfrac{1}{3}$

Giải thích các bước giải:

 $1)\lim(n-\sqrt{n^2+3})$

$=\lim\dfrac{(n-\sqrt{n^2+3}).(n+\sqrt{n^2+3})}{(n+\sqrt{n^2+3})}$

$=\lim\dfrac{n^2-(n^2+3)}{(n+\sqrt{n^2+3})}$

$=\lim\dfrac{-3}{n+\sqrt{n^2+3}}$

$=\lim\dfrac{\dfrac{-3}{n}}{1+\sqrt{1+\dfrac{3}{n^2}}}$

$=\dfrac{0}{2}$

$=0$

$2)\lim(\sqrt[3]{n^2-n^3}+n)$

$=\lim\dfrac{n^2-n^3+n^3}{(\sqrt[3]{n^2-n^3})^2-n\sqrt[3]{n^2-n^3}+n^2}$

$=\lim\dfrac{n^2}{(\sqrt[3]{n^2-n^3})^2-n\sqrt[3]{n^2-n^3}+n^2}$

$=\lim\dfrac{n^2}{(n\sqrt[3]{\dfrac{1}{n}-1})^2-n^2\sqrt[3]{\dfrac{1}{n}-1}+n^2}$

$=\lim\dfrac{n^2}{n^2(\sqrt[3]{\dfrac{1}{n}-1})^2-n^2\sqrt[3]{\dfrac{1}{n}-1}+n^2}$

$=\lim\dfrac{1}{(\sqrt[3]{\dfrac{1}{n}-1})^2-\sqrt[3]{\dfrac{1}{n}-1}+1}$

$=\dfrac{1}{1-(-1)+1}$

$=\dfrac{1}{3}$

Công thức sử dụng :

$a-\sqrt{b}=\dfrac{a^2-b}{a+\sqrt{b}}$

$\sqrt[3]{a}+b=\dfrac{a+b^3}{(\sqrt[3]{a})^2-b\sqrt[3]{a}+b^2}$