1. **Problem:** Find the limit $$\lim_{x \to -\infty} \frac{\sqrt{x^2+5}}{x-\#}$$ where \# is a constant. 2. **Formula and rules:** For limits involving infinity and square roots of quadratic expressions, use the fact that $$\sqrt{x^2 + a} = |x|\sqrt{1 + \frac{a}{x^2}}$$ and as $$x \to -\infty$$, $$|x| = -x$$ because $$x$$ is negative. 3. **Step-by-step solution:** - Rewrite the numerator: $$\sqrt{x^2 + 5} = |x| \sqrt{1 + \frac{5}{x^2}} = -x \sqrt{1 + \frac{5}{x^2}}$$ since $$x \to -\infty$$. - Substitute into the limit: $$\lim_{x \to -\infty} \frac{-x \sqrt{1 + \frac{5}{x^2}}}{x - \#}$$ - Factor $$x$$ from the denominator: $$x - \# = x \left(1 - \frac{\#}{x}\right)$$ - So the limit becomes: $$\lim_{x \to -\infty} \frac{-x \sqrt{1 + \frac{5}{x^2}}}{x \left(1 - \frac{\#}{x}\right)}$$ - Cancel $$x$$: $$\lim_{x \to -\infty} \frac{\cancel{-x} \sqrt{1 + \frac{5}{x^2}}}{\cancel{x} \left(1 - \frac{\#}{x}\right)} = \lim_{x \to -\infty} \frac{-\sqrt{1 + \frac{5}{x^2}}}{1 - \frac{\#}{x}}$$ - As $$x \to -\infty$$, $$\frac{5}{x^2} \to 0$$ and $$\frac{\#}{x} \to 0$$, so: $$= \frac{-\sqrt{1 + 0}}{1 - 0} = \frac{-1}{1} = -1$$ **Final answer:** $$\boxed{-1}$$