Question

Lim x-- dương vô cùng: (4x^2 -3x+1)

Answer 1

\(\lim\limits_{x\rightarrow+\infty}\left(4x^2-3x+1\right)=\lim\limits_{x\rightarrow+\infty}x^2\left(4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)\)

Do \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}x^2=+\infty\\\lim\limits_{x\rightarrow+\infty}\left(4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=4>0\end{matrix}\right.\)

\(\Rightarrow\lim\limits_{x\rightarrow+\infty}x^2\left(4-\dfrac{3}{x}+\dfrac{1}{x^2}\right)=+\infty\)