1. Problem 17: Find $$\lim_{x \to 3} \frac{x}{x-3}$$. Step 1: Substitute $x=3$ directly into the expression: $$\frac{3}{3-3} = \frac{3}{0}$$ which is undefined. Step 2: Analyze the behavior as $x$ approaches 3 from the left ($x \to 3^-$): When $x$ is slightly less than 3, $x-3$ is a small negative number, so the denominator is negative and close to zero. The numerator is close to 3 (positive). Thus, the fraction approaches $-\infty$. Step 3: Analyze the behavior as $x$ approaches 3 from the right ($x \to 3^+$): When $x$ is slightly greater than 3, $x-3$ is a small positive number. The numerator is close to 3 (positive). Thus, the fraction approaches $+\infty$. Step 4: Since the left-hand limit and right-hand limit are not equal, the limit does not exist. --- 2. Problem 18: Find $$\lim_{x \to 2^+} \frac{x}{x^2 - 4}$$. Step 1: Factor the denominator: $$x^2 - 4 = (x-2)(x+2)$$. Step 2: Substitute $x=2$: $$\frac{2}{2^2 - 4} = \frac{2}{0}$$ undefined. Step 3: Analyze the sign of denominator as $x \to 2^+$ (values slightly greater than 2): - $(x-2)$ is a small positive number. - $(x+2)$ is positive (around 4). So denominator is positive small number. Numerator is close to 2 (positive). Step 4: Therefore, the fraction approaches $+\infty$. --- 3. Problem 19: Find $$\lim_{x \to 2^-} \frac{x}{x^2 - 4}$$. Step 1: Using the factorization from before. Step 2: As $x \to 2^-$ (values slightly less than 2): - $(x-2)$ is a small negative number. - $(x+2)$ is positive. So denominator is a small negative number. Numerator is close to 2 (positive). Step 3: The fraction approaches $-\infty$. --- 4. Problem 20: Find $$\lim_{x \to 2} \frac{x}{x^2 - 4}$$. Step 1: Since the left-hand limit is $-\infty$ and the right-hand limit is $+\infty$, the two-sided limit does not exist. --- Final answers: - Problem 17: Limit does not exist. - Problem 18: Limit is $+\infty$. - Problem 19: Limit is $-\infty$. - Problem 20: Limit does not exist.