Evaluate the limit, using L'Hôpital's rule if necessary. (If the limit is infinite, enter ' $$ \infty $$ ' or '- $$ -\infty $$ ', as appropriate. If the limit does not otherwise exist, enter DNE.) $$ \lim _{x \rightarrow 7} \frac{x^{2}-2 x-35}{x-7} $$

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To evaluate the limit:

$$
\lim _{x \rightarrow 7} \frac{x^{2}-2x-35}{x-7}
$$

**Step 1: Direct Substitution**
First, substitute $$ x = 7 $$ into the expression:

$$
\frac{7^2 - 2(7) - 35}{7 - 7} = \frac{49 - 14 - 35}{0} = \frac{0}{0}
$$

This results in an indeterminate form $$ \frac{0}{0} $$, which suggests that we can use L'Hôpital's Rule or factorization to simplify the expression.

**Step 2: Factorization**
Factor the numerator:

$$
x^2 - 2x - 35 = (x - 7)(x + 5)
$$

Now, the expression becomes:

$$
\frac{(x - 7)(x + 5)}{x - 7}
$$

For $$ x \neq 7 $$, we can cancel out the $$ (x - 7) $$ terms:

$$
x + 5
$$

**Step 3: Evaluate the Limit**
Now, substitute $$ x = 7 $$:

$$
7 + 5 = 12
$$

Alternatively, using **L'Hôpital's Rule**, take the derivatives of the numerator and the denominator:

$$
\lim_{x \to 7} \frac{2x - 2}{1} = 2(7) - 2 = 14 - 2 = 12
$$

**Final Answer:**

$$12$$
The limit is 12.

Evaluate the limit, using L’Hôpital’s rule if necessary. (If the limit is infinite, enter ’ ’ or '- ', as appropriate. If the limit does not
otherwise exist, enter DNE.)

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Evaluate the limit, using L’Hôpital’s rule if necessary. (If the limit is infinite, enter ’ ’ or '- ', as appropriate. If the limit does not otherwise exist, enter DNE.)