7) \( \lim _{x \rightarrow-\infty} \frac{\sqrt{x^{2}+x}}{x} \)

To evaluate the limit \( \lim _{x \rightarrow-\infty} \frac{\sqrt{x^{2}+x}}{x} \), we can start by simplifying the expression under the square root: \[ \sqrt{x^{2} + x} = \sqrt{x^{2}(1 + \frac{1}{x})} = |x|\sqrt{1 + \frac{1}{x}} \] As \( x \) approaches \( -\infty \), \( |x| = -x \). Thus, we have: \[ \sqrt{x^{2} + x} = -x \sqrt{1 + \frac{1}{x}} \] Now, substituting this back into our limit, we get: \[ \lim _{x \rightarrow -\infty} \frac{\sqrt{x^{2} + x}}{x} = \lim _{x \rightarrow -\infty} \frac{-x \sqrt{1 + \frac{1}{x}}}{x} \] This simplifies to: \[ \lim _{x \rightarrow -\infty} -\sqrt{1 + \frac{1}{x}} \] As \( x \rightarrow -\infty\), \( \frac{1}{x} \) approaches \( 0 \), so: \[ -\sqrt{1 + 0} = -1 \] Thus, the limit is: \[ \boxed{-1} \]