A)) Choose the correct statement. $$ \lim _{x \rightarrow 6} \frac{1}{(x-6)^{7}}=-\infty $$ $$ \lim _{x \rightarrow 6} \frac{1}{(x-6)^{7}}=\infty $$ $$ \lim _{x \rightarrow 6} \frac{1}{(x-6)^{7}} $$ is neither $$ -\infty $$ nor $$ \infty $$

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To determine the correct statement about the limit

$$
\lim_{x \rightarrow 6} \frac{1}{(x-6)^7},
$$

let's analyze the behavior of the function as $$ x $$ approaches 6 from both sides.

1. **Approaching from the Right ($$ x \rightarrow 6^+ $$):**
   - When $$ x > 6 $$, $$ (x - 6) $$ is a small positive number.
   - Raising a small positive number to the 7th power ($$ (x-6)^7 $$) remains a small positive number.
   - Taking the reciprocal, $$ \frac{1}{(x-6)^7} $$, results in a very large positive number.
   - Therefore, $$ \lim_{x \rightarrow 6^+} \frac{1}{(x-6)^7} = +\infty $$.

2. **Approaching from the Left ($$ x \rightarrow 6^- $$):**
   - When $$ x < 6 $$, $$ (x - 6) $$ is a small negative number.
   - Raising a small negative number to the 7th power ($$ (x-6)^7 $$) remains a small negative number (since 7 is odd).
   - Taking the reciprocal, $$ \frac{1}{(x-6)^7} $$, results in a very large negative number.
   - Therefore, $$ \lim_{x \rightarrow 6^-} \frac{1}{(x-6)^7} = -\infty $$.

Since the left-hand limit ($$ -\infty $$) and the right-hand limit ($$ +\infty $$) are not equal, the two-sided limit does not exist in the traditional sense. Therefore, the correct statement is:

**$$ \lim_{x \rightarrow 6} \frac{1}{(x-6)^7} $$ is neither $$ -\infty $$ nor $$ \infty $$.**

**Answer:**  
$$ \lim _{x \rightarrow 6} \frac{1}{(x-6)^{7}} $$ is neither $$ -\infty $$ nor $$ \infty $$
The limit $$ \lim_{x \rightarrow 6} \frac{1}{(x-6)^7} $$ is neither $$ -\infty $$ nor $$ \infty $$.

A)) Choose the correct statement.

is neither nor

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