Limit at infinity using L'Hopital's Rule

As $x \to \infty$, both $2x^3 - x + 1 \to \infty$ and $x^3 + 4 \to \infty$, so the limit has the indeterminate form $\infty/\infty$, and L'Hopital's Rule applies. Differentiating numerator and denominator separately gives $\lim_{x\to\infty} \frac{6x^2 - 1}{3x^2}.$

This is no longer an indeterminate form. We can evaluate it directly by factoring out $x^2$ from numerator and denominator (or dividing both by $x^2$): $\lim_{x\to\infty} \frac{6x^2 - 1}{3x^2} = \lim_{x\to\infty} \frac{x^2(6 - 1/x^2)}{x^2 \cdot 3} = \lim_{x\to\infty} \frac{6 - 1/x^2}{3} = \frac{6 - 0}{3} = 2.$ Thus the original limit equals $2$.