Question

[títtt]

1) tính \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)

2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

1: \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)\)

\(=\lim\limits_{n\rightarrow\infty}\left[n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^3}+\dfrac{4}{n^5}\right)\right]\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n^5=+\infty\\\lim\limits_{n\rightarrow\infty}\left(-2+\dfrac{4}{n}-\dfrac{3}{n^3}+\dfrac{4}{n^5}\right)=-2< 0\end{matrix}\right.\)

2: \(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^2+2}{n-2}\)

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

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

\(=-\infty\) vì \(\lim\limits_{n\rightarrow\infty}n=+\infty;\lim\limits_{n\rightarrow\infty}\dfrac{-3+\dfrac{2}{n^2}}{1-\dfrac{2}{n}}=-\dfrac{3}{1}=-3< 0\)