1. **State the problem:** Find the limit of the function $$f(x) = \frac{x^2 - 3x}{x^2 - 9}$$ as $$x$$ approaches 3 and -3 by evaluating the function at values close to these points. 2. **Recall the formula and important rules:** The limit $$\lim_{x \to a} f(x)$$ is the value that $$f(x)$$ approaches as $$x$$ gets arbitrarily close to $$a$$. 3. **Simplify the function:** Factor numerator and denominator: $$x^2 - 3x = x(x - 3)$$ $$x^2 - 9 = (x - 3)(x + 3)$$ So, $$f(x) = \frac{x(x - 3)}{(x - 3)(x + 3)}$$ For $$x \neq 3$$, cancel $$x - 3$$: $$f(x) = \frac{x}{x + 3}$$ 4. **Evaluate the limit as $$x \to 3$$:** Substitute $$x = 3$$ into simplified function: $$f(3) = \frac{3}{3 + 3} = \frac{3}{6} = 0.5$$ 5. **Evaluate the limit as $$x \to -3$$:** Simplified function is $$f(x) = \frac{x}{x + 3}$$, but at $$x = -3$$ denominator is zero, so check values near -3: - For $$x$$ approaching -3 from the right (e.g., -2.9999), denominator $$x + 3$$ is positive but very small, numerator $$x$$ is negative, so $$f(x)$$ tends to $$-\infty$$. - For $$x$$ approaching -3 from the left (e.g., -3.0001), denominator $$x + 3$$ is negative and very small, numerator $$x$$ is negative, so $$f(x)$$ tends to $$+\infty$$. 6. **Conclusion:** - $$\lim_{x \to 3} f(x) = 0.5$$ - $$\lim_{x \to -3} f(x)$$ does not exist because the left and right limits are not equal (one tends to $$+\infty$$, the other to $$-\infty$$). 7. **Verification by numerical substitution:** - At $$x=3.1$$, $$f(3.1) \approx \frac{3.1}{6.1} = 0.508197$$ - At $$x=2.9$$, $$f(2.9) \approx \frac{2.9}{5.9} = 0.491525$$ Values approach 0.5 as $$x \to 3$$. - At $$x=-2.9999$$, $$f(-2.9999) \approx \frac{-2.9999}{0.0001} = -29999$$ (large negative) - At $$x=-3.0001$$, $$f(-3.0001) \approx \frac{-3.0001}{-0.0001} = 30001$$ (large positive) This confirms the limit behavior at $$x = -3$$. **Final answers:** $$\lim_{x \to 3} \frac{x^2 - 3x}{x^2 - 9} = 0.5$$ $$\lim_{x \to -3} \frac{x^2 - 3x}{x^2 - 9} \text{ does not exist}$$