Answer
0
Work Step by Step
Consider: $\lim\limits_{x \to \infty}f(x)=\lim\limits_{x \to \infty}\dfrac{2x^2 +3x}{x^3+x+1}$
Now, $f(\infty)=\dfrac{\infty}{\infty}$
The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
That is,
$\lim\limits_{x \to \infty}\dfrac{4x+3}{3x^2+1}=\dfrac{\infty}{\infty}$
Now, again apply L-Hospital's rule:
$\lim\limits_{x \to \infty}\dfrac{4}{6x}=\dfrac{4}{6(\infty)}=0$