Limits at Infinity, Part I — Question 6

Question 6 - Solution

We are given: $$\lim_{x \to \infty} \frac{3x^5 - 2x^2 + 1}{x^5 + 4x^3 - x}$$

Step 1: Divide numerator and denominator by the highest power of $x$, which is $x^5$: $$= \lim_{x \to \infty} \frac{3 - \frac{2}{x^3} + \frac{1}{x^5}}{1 + \frac{4}{x^2} - \frac{1}{x^4}}$$

Step 2: As $x \to \infty$, all terms with negative powers of $x$ go to zero: $$= \frac{3 - 0 + 0}{1 + 0 - 0} = \boxed{3}$$