Let's evaluate the limit: \[ \lim_{x \to \infty} \frac{8x}{\sqrt{x^2+1}} \] **Step 1: Simplify the Expression** Divide numerator and denominator by \( x \): \[ \frac{8x}{\sqrt{x^2+1}} = \frac{8x}{\sqrt{x^2(1+\frac{1}{x^2})}} = \frac{8x}{|x|\sqrt{1+\frac{1}{x^2}}} \] For \( x \to \infty \), \( |x| = x \): \[ = \frac{8x}{x\sqrt{1+\frac{1}{x^2}}} = \frac{8}{\sqrt{1+\frac{1}{x^2}}} \] **Explanation:** After simplification, as \( x \to \infty \), \(\frac{1}{x^2} \to 0\). The expression inside the square root approaches 1. --- **Step 2: Evaluate the Limit** \[ \lim_{x \to \infty} \frac{8}{\sqrt{1+\frac{1}{x^2}}} = \frac{8}{\sqrt{1+0}} = \frac{8}{1} = 8 \] **Explanation:** As \( x \to \infty \), the denominator approaches 1, so the limit is 8. --- **Final Answer:** \[ \boxed{8} \]