Let $$ f(x)=\frac{x^{2}-2 x-48}{x-8} $$. Find a) $$ \lim _{x \rightarrow 8} f(x) $$, b) $$ \operatorname{limim}_{x \rightarrow 0} f(x) $$, and $$ \left.c\right) \lim _{x \rightarrow-6} f(x) $$. a) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. $$ \lim _{x \rightarrow 8} f(x)=\square $$ (Simplify your answer.) B. The limit does not exist.

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Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow 0}\left(\frac{x^{2}-2x-48}{x-8}\right)$$
- step1: Rewrite the expression:
  $$\frac{\lim _{x\rightarrow 0}\left(x^{2}-2x-48\right)}{\lim _{x\rightarrow 0}\left(x-8\right)}$$
- step2: Calculate:
  $$\frac{-48}{\lim _{x\rightarrow 0}\left(x-8\right)}$$
- step3: Calculate:
  $$\frac{-48}{-8}$$
- step4: Calculate:
  $$6$$
Calculate the limit $$ \lim_{x\rightarrow 8} \frac{x^{2}-2 x-48}{x-8} $$.
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
  $$\lim _{x\rightarrow 8}\left(\frac{x^{2}-2x-48}{x-8}\right)$$
- step1: Reduce the fraction:
  $$\lim _{x\rightarrow 8}\left(x+6\right)$$
- step2: Rewrite the expression:
  $$\lim _{x\rightarrow 8}\left(x\right)+\lim _{x\rightarrow 8}\left(6\right)$$
- step3: Calculate:
  $$8+6$$
- step4: Calculate:
  $$14$$
Calculate the limit $$ \lim_{x\rightarrow -6} \frac{x^{2}-2 x-48}{x-8} $$.
Evaluate the limit by following steps:
- step0: Evaluate the limit:
  $$\lim _{x\rightarrow -6}\left(\frac{x^{2}-2x-48}{x-8}\right)$$
- step1: Rewrite the expression:
  $$\frac{\lim _{x\rightarrow -6}\left(x^{2}-2x-48\right)}{\lim _{x\rightarrow -6}\left(x-8\right)}$$
- step2: Calculate:
  $$\frac{0}{\lim _{x\rightarrow -6}\left(x-8\right)}$$
- step3: Calculate:
  $$\frac{0}{-14}$$
- step4: Calculate:
  $$0$$
Let's solve the limits step by step.

### a) $$ \lim _{x \rightarrow 8} f(x) $$

To find this limit, we first substitute $$ x = 8 $$ into the function:

$$
f(8) = \frac{8^{2} - 2 \cdot 8 - 48}{8 - 8} = \frac{64 - 16 - 48}{0} = \frac{0}{0}
$$

This is an indeterminate form, so we need to simplify the function. We can factor the numerator:

$$
x^{2} - 2x - 48 = (x - 8)(x + 6)
$$

Thus, we can rewrite $$ f(x) $$:

$$
f(x) = \frac{(x - 8)(x + 6)}{x - 8}
$$

For $$ x \neq 8 $$, this simplifies to:

$$
f(x) = x + 6
$$

Now we can find the limit as $$ x $$ approaches 8:

$$
\lim_{x \rightarrow 8} f(x) = 8 + 6 = 14
$$

So, the answer is:

A. $$ \lim _{x \rightarrow 8} f(x) = 14 $$

### b) $$ \lim _{x \rightarrow 0} f(x) $$

Now we substitute $$ x = 0 $$:

$$
f(0) = \frac{0^{2} - 2 \cdot 0 - 48}{0 - 8} = \frac{-48}{-8} = 6
$$

So, the limit is:

$$
\lim_{x \rightarrow 0} f(x) = 6
$$

### c) $$ \lim _{x \rightarrow -6} f(x) $$

Now we substitute $$ x = -6 $$:

$$
f(-6) = \frac{(-6)^{2} - 2 \cdot (-6) - 48}{-6 - 8} = \frac{36 + 12 - 48}{-14} = \frac{0}{-14} = 0
$$

So, the limit is:

$$
\lim_{x \rightarrow -6} f(x) = 0
$$

### Summary of Results:
- a) $$ \lim _{x \rightarrow 8} f(x) = 14 $$
- b) $$ \lim _{x \rightarrow 0} f(x) = 6 $$
- c) $$ \lim _{x \rightarrow -6} f(x) = 0 $$
a) $$ \lim _{x \rightarrow 8} f(x) = 14 $$ b) $$ \lim _{x \rightarrow 0} f(x) = 6 $$ c) $$ \lim _{x \rightarrow -6} f(x) = 0 $$

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