2.
\(\lim\limits_{x\rightarrow-\infty}\dfrac{-4x^3+x^2-5}{3x^2+1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3\left(-4+\dfrac{1}{x}-\dfrac{5}{x^3}\right)}{x^2\left(3+\dfrac{1}{x^2}\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}x.\left(\dfrac{-4+\dfrac{1}{x}-\dfrac{5}{x^3}}{3+\dfrac{1}{x^2}}\right)\)
Do \(\lim\limits_{x\rightarrow-\infty}x=-\infty\)
\(\lim\limits_{x\rightarrow-\infty}\left(\dfrac{-4+\dfrac{1}{x}-\dfrac{5}{x^3}}{3+\dfrac{1}{x^2}}\right)=-\dfrac{4}{3}< 0\)
\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x\left(\dfrac{-4+\dfrac{1}{x}-\dfrac{5}{x^3}}{3+\dfrac{1}{x^2}}\right)=+\infty\)
4.
Ta có: \(\lim\limits_{x\rightarrow2^+}\left(x^2+x+4\right)=2^2+2+4=10>0\)
\(\lim\limits_{x\rightarrow2^+}\left(x-2\right)=0\)
Và: \(x-2>0\) khi \(x>2\)
\(\Rightarrow\lim\limits_{x\rightarrow2^+}\dfrac{x^2+x+4}{x-2}=+\infty\)
