\(\lim\limits_{x\to \infty} \dfrac{x^4+3x^3-2x+5}{3x^4+5x^2-13x} \)
When determining limits at infinity, only the highest power terms in the numerator and denominator are relevant.
$$\lim\limits_{x\to \infty} \dfrac{x^4+3x^3-2x+5}{3x^4+5x^2-13x} \approx \lim\limits_{x\to \infty} \dfrac{x^4}{3x^4} $$
$$ = \lim\limits_{x\to \infty} \dfrac{1}{3} $$
$$ = \boxed{\frac{1}{3}} $$
We can divide the numerator and denominator by \(x^4\).
Numerator
$$ x^4+3x^3-2x+5 \div x^4 $$
$$ = \frac{x^4+3x^3-2x+5}{x^4} $$
$$ = 1+\frac{3}{x}-\frac{2}{x^3}+\frac{5}{x^4} $$
Denominator
$$ 3x^4+5x^2-13x \div x^4 $$
$$ = \frac{3x^4+5x^2-13x}{x^4} $$
$$ = 3+\frac{5}{x^2}-\frac{13}{x^3} $$
$$ \lim\limits_{x\to\infty} \frac{1+\frac{3}{x}-\frac{2}{x^3}+\frac{5}{x^4}}{3+\frac{5}{x^2}-\frac{13}{x^3}} $$
Recall that as \(x\) approaches large values, fractions in the form of \(\dfrac{c}{x^r}\) approach 0.
$$ = \frac{1+0+0+0}{3+0+0} $$
$$ = \boxed{\frac{1}{3}} $$