6x \( \quad \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+x}}{x} \)

To evaluate the limit \( \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+x}}{x} \), we can start by simplifying the expression inside the limit. We can factor \( x^2 \) out of the square root: \[ \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+x}}{x} = \lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}(1 + \frac{1}{x})}}{x} \] This simplifies to: \[ = \lim _{x \rightarrow \infty} \frac{x \sqrt{1 + \frac{1}{x}}}{x} \] The \( x \) in the numerator and the denominator cancel each other out: \[ = \lim _{x \rightarrow \infty} \sqrt{1 + \frac{1}{x}} \] As \( x \) approaches infinity, \( \frac{1}{x} \) approaches 0. Therefore, we have: \[ = \sqrt{1 + 0} = \sqrt{1} = 1 \] So, the limit evaluates to 1. Then, multiplying by 6 gives: \[ 6 \times 1 = 6. \] Thus, the final answer is: \[ \boxed{6}. \]