10. [-/1 Points] DETAILS MY NOTES SCALCET9 2.XP.6.018. Find the limit. (If the limit is infinite, enter ' \( \infty \) ' or '- \( \infty \) ', as appropriate. If the limit does not otherwise exist, enter DNE.) \( \lim _{x \rightarrow \infty}\left(\sqrt{16 x^{2}+x}-4 x\right) \) ASK YOUR TEACHER

To find the limit as \( x \) approaches infinity for the expression \( \sqrt{16x^2 + x} - 4x \), we can simplify it. Start by factoring out \( x^2 \) from the square root: \[ \sqrt{16x^2 + x} = \sqrt{x^2(16 + \frac{1}{x})} = x\sqrt{16 + \frac{1}{x}} \] Therefore, our limit can be rewritten as: \[ \lim_{x \rightarrow \infty} (x\sqrt{16 + \frac{1}{x}} - 4x) = \lim_{x \rightarrow \infty} x(\sqrt{16 + \frac{1}{x}} - 4) \] Now, as \( x \) approaches infinity, \( \frac{1}{x} \) approaches \( 0 \), leading to: \[ \sqrt{16 + 0} - 4 = 0 \] Thus, the expression simplifies to \( x \cdot 0 \), suggesting that we need to look deeper into how quickly the square root approaches \( 4 \). To do this, we can express \( \sqrt{16 + \frac{1}{x}} \) as: \[ 4 + \frac{1}{2(4)}(\frac{1}{x}) + O\left(\frac{1}{x^2}\right) = 4 + \frac{1}{8x} + O\left(\frac{1}{x^2}\right) \] Now substituting this back gives: \[ x\left( \left(4 + \frac{1}{8x}\right) - 4 \right) = x\left(\frac{1}{8x}\right) = \frac{1}{8} \] Thus, the limit is: \[ \lim_{x \to \infty} \left(\sqrt{16x^2+x} - 4x\right) = \frac{1}{8} \] So, the final answer is \( \frac{1}{8} \).