1. **State the problem:** Find the limits of the function $$f(x) = \sqrt{x^2 + 5x - x}$$ as $$x \to \infty$$ and $$x \to -\infty$$. 2. **Simplify the function inside the square root:** $$x^2 + 5x - x = x^2 + 4x$$ So, $$f(x) = \sqrt{x^2 + 4x}$$. 3. **Factor inside the square root to analyze behavior:** $$f(x) = \sqrt{x^2 + 4x} = \sqrt{x^2(1 + \frac{4}{x})} = \sqrt{x^2} \sqrt{1 + \frac{4}{x}}$$. 4. **Recall that $$\sqrt{x^2} = |x|$$, so:** $$f(x) = |x| \sqrt{1 + \frac{4}{x}}$$. 5. **Analyze the limit as $$x \to \infty$$:** - When $$x \to \infty$$, $$|x| = x$$. - Also, $$\lim_{x \to \infty} \sqrt{1 + \frac{4}{x}} = \sqrt{1 + 0} = 1$$. Therefore, $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} x \cdot 1 = \infty$$. 6. **Analyze the limit as $$x \to -\infty$$:** - When $$x \to -\infty$$, $$|x| = -x$$ (since $$x$$ is negative, absolute value is $$-x$$). - Also, $$\lim_{x \to -\infty} \sqrt{1 + \frac{4}{x}} = 1$$. Therefore, $$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} (-x) \cdot 1 = \infty$$. 7. **Final answer:** $$\lim_{x \to \infty} f(x) = \infty$$ and $$\lim_{x \to -\infty} f(x) = \infty$$. This matches the first option given.