1. Find the limits of each of the following functions as \( x \) approaches -3 . \( \begin{array}{l}\text { (a) } f(x)=x^{3}-2 x^{2}+3 \\ \text { (b) } f(x)=\frac{5+x^{2}}{x-3} \\ \text { (c) } f(x)=\frac{(x+3)(x-5)}{(x-3)(x+5)}\end{array} \)

To find the limits of each function as \( x \) approaches -3, let's evaluate them one by one. **(a)** For \( f(x) = x^3 - 2x^2 + 3 \): To find the limit, simply substitute \( x = -3 \): \[ f(-3) = (-3)^3 - 2(-3)^2 + 3 = -27 - 18 + 3 = -42 \] Thus, \( \lim_{x \to -3} f(x) = -42 \). **(b)** For \( f(x) = \frac{5 + x^2}{x - 3} \): Substituting \( x = -3 \) gives us: \[ f(-3) = \frac{5 + (-3)^2}{-3 - 3} = \frac{5 + 9}{-6} = \frac{14}{-6} = -\frac{7}{3} \] Thus, \( \lim_{x \to -3} f(x) = -\frac{7}{3} \). **(c)** For \( f(x) = \frac{(x+3)(x-5)}{(x-3)(x+5)} \): Substituting \( x = -3 \): \[ f(-3) = \frac{(-3+3)(-3-5)}{(-3-3)(-3+5)} = \frac{0 \cdot (-8)}{(-6)(2)} = \frac{0}{-12} = 0 \] Thus, \( \lim_{x \to -3} f(x) = 0 \). In summary: - (a) \( \lim_{x \to -3} f(x) = -42 \) - (b) \( \lim_{x \to -3} f(x) = -\frac{7}{3} \) - (c) \( \lim_{x \to -3} f(x) = 0 \)