Answer
$\dfrac{5}{7}$
Work Step by Step
Consider: $\lim\limits_{x \to \infty}f(x)=\lim\limits_{x \to \infty}\dfrac{5x^3-2x}{7x^3+3}$
Now, $f(\infty)=\dfrac{\infty}{\infty}$
The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
Thus:
$\lim\limits_{x \to \infty}\dfrac{15x^2-2}{21x^2}=\dfrac{\infty}{\infty}$
We need to again apply L-Hospital's rule.
$\lim\limits_{x \to \infty}\dfrac{30x}{42x}=\dfrac{\infty}{\infty}$
Now, again apply L-Hospital's rule.
Thus, $\lim\limits_{x \to \infty}\dfrac{30}{42}=\dfrac{5}{7}$
