Let `L_(1)=lim_(xrarr4) (x-6)^(x)and L_(2)=lim_(xrarr4) (x-6)^(4)`.
Which of the following is true?

Options

A. Both `L_(1) and L_(2)` exists
B. Neither `L_(1)` nor `L_(2)` exists
C. `L_(1)` exists but `L_(2)` does not exist
D. `L_(2)` exists but `L_(1)` does not exist

Explanation

To solve the limits \( L_1 = \lim_{x \to 4} (x - 6)^x \) and \( L_2 = \lim_{x \to 4} (x - 6)^4 \), we will analyze each limit step by step.

### Step 1: Evaluate \( L_1 = \lim_{x \to 4} (x - 6)^x \)

1. Substitute \( x = 4 \) into the expression:
\[ L_1 = (4 - 6)^4 = (-2)^4 \]
However, we notice that as \( x \) approaches 4, the base \( (x - 6) \) approaches \( -2 \), while the exponent \( x \) approaches \( 4 \).

2. Since the base \( (x - 6) \) approaches a negative number and the exponent \( x \) is variable, we have a situation where we are raising a negative number to a variable power. This can lead to undefined behavior depending on the value of \( x \).

3. Therefore, we conclude that:
\[ L_1 \text{ does not exist.} \]

### Step 2: Evaluate \( L_2 = \lim_{x \to 4} (x - 6)^4 \)

1. Substitute \( x = 4 \) into the expression:
\[ L_2 = (4 - 6)^4 = (-2)^4 \]

2. Since \( 4 \) is a constant exponent, we can compute this directly:
\[ L_2 = 16 \]

### Conclusion

- \( L_1 \) does not exist because we have a variable exponent with a negative base.
- \( L_2 \) exists and equals \( 16 \).

Thus, the correct statement is that \( L_1 \) does not exist and \( L_2 \) exists.

### Final Answer

- \( L_1 \text{ does not exist} \)
- \( L_2 = 16 \)