Let \( f(x)=\frac{x^{2}-4 x-32}{x-8} \). Find a) \( \lim _{x \rightarrow 8} f(x) \), b) \( \lim _{x \rightarrow 0} f(x) \), and c) \( \lim _{x \rightarrow-4} f(x) \)

To find the limits of the function \( f(x) = \frac{x^{2} - 4x - 32}{x - 8} \) at the specified points, we'll first simplify the function by factoring the numerator. ### Step 1: Factor the Numerator The numerator is a quadratic expression: \[ x^{2} - 4x - 32 \] To factor this, we look for two numbers that multiply to \(-32\) and add to \(-4\). The numbers \(-8\) and \(4\) satisfy this: \[ x^{2} - 4x - 32 = (x - 8)(x + 4) \] So, the function becomes: \[ f(x) = \frac{(x - 8)(x + 4)}{x - 8} \] For \( x \neq 8 \), we can cancel out the \((x - 8)\) terms: \[ f(x) = x + 4 \quad \text{for} \quad x \neq 8 \] ### Step 2: Compute the Limits **a) \( \lim_{x \rightarrow 8} f(x) \)** Since \( f(x) = x + 4 \) for \( x \neq 8 \), as \( x \) approaches 8: \[ \lim_{x \rightarrow 8} f(x) = 8 + 4 = 12 \] **b) \( \lim_{x \rightarrow 0} f(x) \)** Similarly, as \( x \) approaches 0: \[ \lim_{x \rightarrow 0} f(x) = 0 + 4 = 4 \] **c) \( \lim_{x \rightarrow -4} f(x) \)** As \( x \) approaches -4: \[ \lim_{x \rightarrow -4} f(x) = -4 + 4 = 0 \] ### Final Answers - **a)** \( \lim_{x \rightarrow 8} f(x) = 12 \) - **b)** \( \lim_{x \rightarrow 0} f(x) = 4 \) - **c)** \( \lim_{x \rightarrow -4} f(x) = 0 \) **Answer:** a) 12; b) 4; c) 0.