Limits at Infinity, Part I — Question 8

Question 8 - Solution

We are given: $$\lim_{x \to \infty} \frac{6x^2 - x + 8}{\sqrt{x^4 + 3x^2 + 1}}$$

Step 1: Rewrite the denominator using properties of radicals: $$\sqrt{x^4 + 3x^2 + 1} = x^2 \sqrt{1 + \frac{3}{x^2} + \frac{1}{x^4}}$$

So the entire expression becomes: $$\frac{6x^2 - x + 8}{x^2 \sqrt{1 + \frac{3}{x^2} + \frac{1}{x^4}}} = \frac{x^2\left(6 - \frac{1}{x} + \frac{8}{x^2}\right)}{x^2 \sqrt{1 + \frac{3}{x^2} + \frac{1}{x^4}}}$$

Cancel $x^2$: $$= \frac{6 - \frac{1}{x} + \frac{8}{x^2}}{\sqrt{1 + \frac{3}{x^2} + \frac{1}{x^4}}}$$

Step 2: Let $x \to \infty$: $$= \frac{6 - 0 + 0}{\sqrt{1 + 0 + 0}} = \frac{6}{1} = \boxed{6}$$