Đáp án:
a) \(\lim (\sqrt{n^{2}-1}-2n)=-\infty\)
b) \(\lim (\sqrt{n^{2}-3n}-n)=\dfrac{-3}{2}\)
c) \(\lim (\sqrt{n^{2}-n-3}-n)=\dfrac{-1}{2}\)
d) $\lim(\sqrt{n^2+2n}-\sqrt{n^2+2})=1$
e) $\lim(\sqrt{n^2+n+2}-\sqrt{n^2-1})=\dfrac{1}{2}$
f) $\lim(\sqrt{n^2-n}-n)=\dfrac{-1}{2}$
Giải thích các bước giải:
a) \(\lim (\sqrt{n^{2}-1}-2n)=\lim \left[{n\left({\sqrt{1-\dfrac{1}{n^{2}}}-2}\right)}\right]\)
\(=\lim [n(1-2)]=\lim[n(-1)]=-\infty\)
b) \(\lim (\sqrt{n^{2}-3n}-n)= \lim\dfrac{n^2-3n-n^2}{\sqrt{n^{2}-3n}+n}\)
$=\lim\dfrac{-3n}{\sqrt{n^{2}-3n}+n}=\lim\dfrac{-3}{\sqrt{1-\dfrac{3}{n}}+1}$
$=\dfrac{-3}{1+1}=\dfrac{-3}{2}$
c) \(\lim (\sqrt{n^{2}-n-3}-n)=\lim\dfrac{n^2-n-3-n^2}{\sqrt{n^{2}-n-3}+n}\)
$=\lim\dfrac{-n-3}{\sqrt{n^{2}-n-3}+n}=\lim\dfrac{-1-\dfrac{3}{n}}{\sqrt{1-\dfrac{1}{n}-\dfrac{3}{n^2}}+1}$
$=\dfrac{-1}{1+1}=\dfrac{-1}{2}$
d) $\lim(\sqrt{n^2+2n}-\sqrt{n^2+2})=\lim\dfrac{n^2+2n-(n^2+2)}{\sqrt{n^2+2n}+\sqrt{n^2+2}}$
$=\lim\dfrac{2n-2}{\sqrt{n^2+2n}+\sqrt{n^2+2}}=\lim\dfrac{2-\dfrac{2}{n}}{\sqrt{1+\dfrac{2}{n}}+\sqrt{1+\dfrac{2}{n^2}}}=\dfrac{2}{1+1}=1$
e) $\lim(\sqrt{n^2+n+2}-\sqrt{n^2-1})=\lim\dfrac{n^2+n+2-(n^2-1)}{\sqrt{n^2+n+2}+\sqrt{n^2-1}}$
$\lim\dfrac{n+3}{\sqrt{n^2+n+2}+\sqrt{n^2-1}}=\lim\dfrac{1+\dfrac{3}{n}}{\sqrt{1+\dfrac{1}{n}+\dfrac{2}{n^2}}+\sqrt{1-\dfrac{1}{n}}}$
$=\dfrac{1}{1+1}=\dfrac{1}{2}$
f) $\lim(\sqrt{n^2-n}-n)=\lim\dfrac{n^2-n-n^2}{\sqrt{n^2-n}+n}=\lim\dfrac{-n}{\sqrt{n^2-n}+n}$
$=\lim\dfrac{-1}{\sqrt{1-\dfrac{1}{n}}+1}=\dfrac{-1}{1+1}=\dfrac{-1}{2}$
