Determine if the function \( f(x) = \frac{2x + 1}{x^2 - 4} \) is continuous at \( x = 2 \).

To determine if the function \( f(x) = \frac{2x + 1}{x^2 - 4} \) is continuous at \( x = 2 \), we need to check the following three conditions: 1. \( f(2) \) is defined. 2. The limit \( \lim_{x \to 2} f(x) \) exists. 3. The limit \( \lim_{x \to 2} f(x) \) is equal to \( f(2) \). Let's start by evaluating \( f(2) \): \[ f(2) = \frac{2(2) + 1}{2^2 - 4} = \frac{4 + 1}{4 - 4} = \frac{5}{0} \] Since the denominator is zero, \( f(2) \) is undefined. Therefore, the first condition is not satisfied. Next, we can also check the limit as \( x \) approaches 2: \[ \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{2x + 1}{x^2 - 4} \] The denominator can be factored: \[ x^2 - 4 = (x - 2)(x + 2) \] Thus, we have: \[ f(x) = \frac{2x + 1}{(x - 2)(x + 2)} \] As \( x \) approaches 2, the denominator approaches zero, and we need to check the behavior of the limit: \[ \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{2x + 1}{(x - 2)(x + 2)} \] Substituting \( x = 2 \) into the numerator gives \( 2(2) + 1 = 5 \), and the denominator approaches zero. Therefore, the limit does not exist because the function approaches infinity (positive or negative depending on the direction of approach). Since \( f(2) \) is undefined and the limit does not exist, we conclude that the function \( f(x) \) is not continuous at \( x = 2 \). Thus, the final answer is: \[ \text{The function } f(x) \text{ is not continuous at } x = 2. \]