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

Question

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

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**(a) Evaluating $$ f(x)=x^3-2x^2+3 $$ as $$ x\to -3 $$:**

Since polynomial functions are continuous everywhere, we substitute $$ x=-3 $$:

$$
f(-3)=(-3)^3 - 2(-3)^2+3 = -27 - 2(9) + 3 = -27 -18 +3 = -42.
$$

Thus, the limit is $$\boxed{-42}$$.

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**(b) Evaluating $$ f(x)=\frac{5+x^2}{x-3} $$ as $$ x\to -3 $$:**

The function is continuous at $$ x=-3 $$ because the denominator is not zero. Substitute $$ x=-3 $$:

$$
f(-3)= \frac{5+(-3)^2}{-3-3} = \frac{5+9}{-6} = \frac{14}{-6} = -\frac{7}{3}.
$$

Thus, the limit is $$\boxed{-\frac{7}{3}}$$.

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**(c) Evaluating $$ f(x)=\frac{(x+3)(x-5)}{(x-3)(x+5)} $$ as $$ x\to -3 $$:**

The function is defined for $$ x \neq 3 $$ and $$ x \neq -5 $$. 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, the limit is $$\boxed{0}$$.
- The limit of $$ f(x)=x^3-2x^2+3 $$ as $$ x $$ approaches -3 is **-42**. - The limit of $$ f(x)=\frac{5+x^2}{x-3} $$ as $$ x $$ approaches -3 is **-7⁄3**. - The limit of $$ f(x)=\frac{(x+3)(x-5)}{(x-3)(x+5)} $$ as $$ x $$ approaches -3 is **0**.

Find the limits of each of the following functions as approaches -3 .

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