Let \( f \) be the function defined by \( f(x)=\frac{(x-2)\left(x^{2}-k^{2}\right)}{\left(x^{2}-4\right)(x-k)} \), where \( k \) is a constant. For what value of \( k \), if any, does \( \lim _{x \rightarrow 2} f(x)=3 \) ?

To find the value of \( k \) such that \( \lim_{x \to 2} f(x) = 3 \), we start by analyzing the function: \[ f(x) = \frac{(x-2)(x^2 - k^2)}{(x^2 - 4)(x - k)} \] First, we need to evaluate the limit as \( x \) approaches 2. We notice that both the numerator and denominator will approach 0 when \( x = 2 \): - The numerator: \[ (2-2)(2^2 - k^2) = 0 \] - The denominator: \[ (2^2 - 4)(2 - k) = (4 - 4)(2 - k) = 0 \] Since both the numerator and denominator approach 0, we can apply L'Hôpital's Rule or simplify the expression. We will simplify the expression first. The denominator can be factored as follows: \[ x^2 - 4 = (x-2)(x+2) \] Thus, we can rewrite \( f(x) \): \[ f(x) = \frac{(x-2)(x^2 - k^2)}{(x-2)(x+2)(x-k)} \] For \( x \neq 2 \), we can cancel \( (x-2) \): \[ f(x) = \frac{x^2 - k^2}{(x+2)(x-k)} \] Next, we can substitute \( x = 2 \) into the simplified function: \[ f(2) = \frac{2^2 - k^2}{(2+2)(2-k)} = \frac{4 - k^2}{4(2-k)} \] Now, we need to find the limit as \( x \) approaches 2: \[ \lim_{x \to 2} f(x) = \frac{4 - k^2}{4(2-k)} \] We want this limit to equal 3: \[ \frac{4 - k^2}{4(2-k)} = 3 \] Cross-multiplying gives: \[ 4 - k^2 = 12(2 - k) \] Expanding the right side: \[ 4 - k^2 = 24 - 12k \] Rearranging the equation: \[ k^2 - 12k + 20 = 0 \] Now we can solve this quadratic equation using the quadratic formula: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 20}}{2 \cdot 1} \] Calculating the discriminant: \[ 144 - 80 = 64 \] Thus, we have: \[ k = \frac{12 \pm 8}{2} \] Calculating the two possible values for \( k \): 1. \( k = \frac{20}{2} = 10 \) 2. \( k = \frac{4}{2} = 2 \) We have two potential values for \( k \): \( k = 10 \) and \( k = 2 \). However, we need to check if both values are valid in the context of the limit. If \( k = 2 \), the original function \( f(x) \) becomes undefined at \( x = 2 \) because the denominator \( (x - k) \) becomes zero. Therefore, \( k = 2 \) is not a valid solution. Thus, the only valid value for \( k \) that satisfies \( \lim_{x \to 2} f(x) = 3 \) is: \[ \boxed{10} \]