Đáp án D sai

Hàm đa thức có giới hạn tại mọi điểm và tại tất cả các điểm thì giới hạn trái luôn bằng giới hạn phải

\(a,lim\dfrac{-3n^3+1}{2n+5}\)

\(=lim\dfrac{-3+\dfrac{1}{n^3}}{2n^2+\dfrac{5}{n^3}}=\dfrac{-3}{2n^2}=\dfrac{1}{n^2}\times\dfrac{-3}{2}=\)-∞

\(\lim\dfrac{n^3-2n+1}{-3n-4}=\lim\dfrac{n^2-2+\dfrac{1}{n}}{-3-\dfrac{4}{n}}=\dfrac{+\infty}{-3}=-\infty\)

\(\lim\left(3n-\sqrt{9n^2+1}\right)=\lim\dfrac{-1}{3n+\sqrt{9n^2+1}}=\lim\dfrac{-\dfrac{1}{n}}{3+\sqrt{9+\dfrac{1}{n^2}}}=\dfrac{0}{3+3}=0\)

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

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

\(a,lim\dfrac{2n+1}{-3n+2}\)

\(=lim\dfrac{2+\dfrac{1}{n}}{-3+\dfrac{2}{n}}=-\dfrac{2}{3}\)

\(b,lim\dfrac{5n^3-2n+1}{n-2n^3}\)

\(=lim\dfrac{5-\dfrac{2}{n^2}+\dfrac{1}{n^3}}{\dfrac{1}{n^2}-2}=\dfrac{5}{-2}\)

\(\lim\dfrac{\sqrt{n^2+n-1}-n}{2n+3}=\lim\dfrac{n-1}{\left(2n+3\right)\left(\sqrt{n^2+n-1}+n\right)}\)

\(=\lim\dfrac{1-\dfrac{1}{n}}{\left(2+\dfrac{3}{n}\right)\left(\sqrt{n^2+n-1}+n\right)}=\dfrac{1}{2.+\infty}=0\)

a. ĐKXĐ: \(n\ne\dfrac{-3}{2}\); \(\left[{}\begin{matrix}x< \dfrac{-1-\sqrt{5}}{2}\\x>\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{n^2+n-1}-n}{2n+3}=\)\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{n}-\dfrac{1}{n^2}}-1}{2+\dfrac{3}{n}}=0\)

\(b,lim\left(^3\sqrt{n^3+1}+\sqrt{n^2+n}-2n\right)\)

\(=limn\left(^3\sqrt{1+\dfrac{1}{n^3}}+\sqrt{1+\dfrac{1}{n}}-2\right)\)

\(=n\left(1+1-2\right)=0\)

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

\(=\lim\left(\dfrac{1}{\sqrt[3]{\left(n^3+1\right)^2}+n\sqrt[3]{n^3+1}+n^2}+\dfrac{1}{\sqrt[]{1+\dfrac{1}{n}}+1}\right)=0+\dfrac{1}{2}=\dfrac{1}{2}\)

\(a,lim\left(\sqrt{n^2+n+1}-n\right)\)

\(=lim\dfrac{n^2+n+1-n^2}{\sqrt{n^2+n+1}+n}\)

\(=lim\dfrac{1+\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)

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

\(=\dfrac{+\infty\left(0-2\right)}{3}=-\infty\)