1. **Problem Statement:** Given the function $$g(x) = \frac{x^2 + 9}{|x - 3|}$$ (a) Determine the domain of $$g(x)$$. (b) Evaluate $$\lim_{x \to 3^+} g(x)$$. (c) Evaluate $$\lim_{x \to 3^-} g(x)$$. (d) Compare the two limits and conclude if the limit of $$g(x)$$ exists at $$x=3$$. 2. **Solution:** **(a) Domain:** - The denominator is $$|x-3|$$ which is zero when $$x=3$$. - Division by zero is undefined. - Therefore, domain is $$\{x \in \mathbb{R} \mid x \neq 3\}$$. **(b) Right-hand limit at $$x=3$$:** - For $$x \to 3^+$$, $$x-3 > 0$$ so $$|x-3| = x-3$$. - The function behaves as $$g(x) = \frac{x^2 + 9}{x-3}$$. - Evaluate:\ Numerator at 3: $$3^2 + 9 = 9+9 = 18$$. - As $$x \to 3^+$$, $$x-3 \to 0^+$$ (small positive number). - So, $$g(x) \to \frac{18}{0^+} = +\infty$$. **(c) Left-hand limit at $$x=3$$:** - For $$x \to 3^-$$, $$x-3 < 0$$ so $$|x-3| = -(x-3) = 3-x$$. - The function behaves as $$g(x) = \frac{x^2 + 9}{3-x}$$. - Numerator at 3 remains 18. - As $$x \to 3^-$$, $$3-x \to 0^+$$. - So, $$g(x) \to \frac{18}{0^+} = +\infty$$. **(d) Compare limits:** - $$\lim_{x \to 3^+} g(x) = +\infty$$ and $$\lim_{x \to 3^-} g(x) = +\infty$$. - Both one-sided limits approach the same infinite value. - Thus, the limit $$\lim_{x \to 3} g(x)$$ does not exist as a finite number but tends to $$+\infty$$. **Final answers:** - Domain: $$\mathbb{R}\setminus \{3\}$$. - $$\lim_{x \to 3^+} g(x) = +\infty$$. - $$\lim_{x \to 3^-} g(x) = +\infty$$. - Limit at $$x=3$$ does not exist finitely but tends to $$+\infty$$.