Question

e) lim\(\dfrac{17}{x^2+1}\)(x-->+\(\infty\))

f) lim\(\dfrac{-2x^2+x-1}{3+x}\)(x-->+\(\infty\))

Answer 1

\(\lim\limits_{x\rightarrow+\infty}\dfrac{17}{x^2+1}=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{17}{x^2}}{1+\dfrac{1}{x^2}}=\dfrac{0}{1}=0\)

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

Do \(\lim\limits_{x\rightarrow+\infty}x=+\infty\)

\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-2+\dfrac{1}{x}-\dfrac{1}{x^2}}{\dfrac{3}{x}+1}\right)=-2< 0\)

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