1. **State the problem:** We want to estimate the limits: (a) $$\lim_{x \to 3} \frac{x^2 - 3x}{x^2 - 9}$$ (b) $$\lim_{x \to -3} \frac{x^2 - 3x}{x^2 - 9}$$ by evaluating the function at values close to 3 and -3 respectively. 2. **Simplify the function:** The function is $$f(x) = \frac{x^2 - 3x}{x^2 - 9}$$. 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 $$x - 3$$: $$f(x) = \frac{x}{x + 3}$$ 3. **Evaluate the function near the limits:** (a) For $$x \to 3$$, use $$f(x) = \frac{x}{x + 3}$$: - At $$x=3.1$$, $$f(3.1) = \frac{3.1}{6.1} \approx 0.508197$$ - At $$x=3.05$$, $$f(3.05) = \frac{3.05}{6.05} \approx 0.504132$$ - At $$x=3.01$$, $$f(3.01) = \frac{3.01}{6.01} \approx 0.500832$$ - At $$x=3.001$$, $$f(3.001) = \frac{3.001}{6.001} \approx 0.500083$$ - At $$x=3.0001$$, $$f(3.0001) = \frac{3.0001}{6.0001} \approx 0.500008$$ - At $$x=2.9$$, $$f(2.9) = \frac{2.9}{5.9} \approx 0.491525$$ - At $$x=2.95$$, $$f(2.95) = \frac{2.95}{5.95} \approx 0.495798$$ - At $$x=2.99$$, $$f(2.99) = \frac{2.99}{5.99} \approx 0.499166$$ - At $$x=2.999$$, $$f(2.999) = \frac{2.999}{5.999} \approx 0.499916$$ - At $$x=2.9999$$, $$f(2.9999) = \frac{2.9999}{5.9999} \approx 0.499992$$ Values approach approximately 0.5. (b) For $$x \to -3$$, use the same simplified function: - At $$x=-2.5$$, $$f(-2.5) = \frac{-2.5}{0.5} = -5.000000$$ - At $$x=-2.9$$, $$f(-2.9) = \frac{-2.9}{0.1} = -29.000000$$ - At $$x=-2.95$$, $$f(-2.95) = \frac{-2.95}{0.05} = -59.000000$$ - At $$x=-2.99$$, $$f(-2.99) = \frac{-2.99}{0.01} = -299.000000$$ - At $$x=-2.999$$, $$f(-2.999) = \frac{-2.999}{0.001} = -2999.000000$$ - At $$x=-2.9999$$, $$f(-2.9999) = \frac{-2.9999}{0.0001} = -29999.000000$$ - At $$x=-3.5$$, $$f(-3.5) = \frac{-3.5}{-0.5} = 7.000000$$ - At $$x=-3.1$$, $$f(-3.1) = \frac{-3.1}{-0.1} = 31.000000$$ - At $$x=-3.05$$, $$f(-3.05) = \frac{-3.05}{-0.05} = 61.000000$$ - At $$x=-3.01$$, $$f(-3.01) = \frac{-3.01}{-0.01} = 301.000000$$ - At $$x=-3.001$$, $$f(-3.001) = \frac{-3.001}{-0.001} = 3001.000000$$ - At $$x=-3.0001$$, $$f(-3.0001) = \frac{-3.0001}{-0.0001} = 30001.000000$$ Values approach $$-\infty$$ from the right side and $$+\infty$$ from the left side, so the limit does not exist. 4. **Conclusion:** (a) $$\lim_{x \to 3} \frac{x^2 - 3x}{x^2 - 9} = 0.5$$ (b) $$\lim_{x \to -3} \frac{x^2 - 3x}{x^2 - 9}$$ does not exist because the left and right limits diverge to $$+\infty$$ and $$-\infty$$ respectively.