9. [-/1 Points] DETAILS MY NOTES SCALCET9 2.6.023. Find the limit. (If the limit is infinite, enter ' \( \infty \) ' or '- \( \infty \) ', as appropriate. If the limit does not otherwise exist, enter DNE.) \[ \lim _{x \rightarrow \infty} \frac{\sqrt{x+6 x^{2}}}{3 x-1} \]

To find the limit as \( x \) approaches infinity for the expression \( \frac{\sqrt{x + 6x^2}}{3x - 1} \), we will simplify the expression first. 1. Start by factoring out \( x^2 \) from the square root in the numerator: \[ \sqrt{x + 6x^2} = \sqrt{x^2(6 + \frac{1}{x})} = x\sqrt{6 + \frac{1}{x}}. \] Thus, the original limit becomes: \[ \lim_{x \to \infty} \frac{x\sqrt{6 + \frac{1}{x}}}{3x - 1}. \] 2. Now simplify the fraction: \[ = \lim_{x \to \infty} \frac{\sqrt{6 + \frac{1}{x}}}{3 - \frac{1}{x}}. \] 3. As \( x \) approaches infinity, \( \frac{1}{x} \) approaches \( 0 \): \[ \sqrt{6 + \frac{1}{x}} \to \sqrt{6} \quad \text{and} \quad 3 - \frac{1}{x} \to 3. \] 4. Thus, we find: \[ = \frac{\sqrt{6}}{3}. \] Putting it all together, we have: \[ \lim _{x \rightarrow \infty} \frac{\sqrt{x+6 x^{2}}}{3 x-1} = \frac{\sqrt{6}}{3}. \] The result is: \(\frac{\sqrt{6}}{3}\).