## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{5}{7}$
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, $\lim\limits_{x \to \infty}\dfrac{30}{42}=\dfrac{5}{7}$