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 problem, we need to evaluate the two limits \( L_1 \) and \( L_2 \).

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

1. Substitute \( x = 4 \):
\[ L_1 = (4 - 6)^4 = (-2)^4 = 16 \]
However, we need to consider the limit as \( x \) approaches 4 from both sides.

2. As \( x \) approaches 4 from the left (i.e., \( x \to 4^- \)), \( x - 6 \) approaches -2, and since the exponent \( x \) is approaching 4 (which is even), the base is negative and the exponent is positive. Thus:
\[ L_1 \text{ does not exist as } x \to 4^- \text{ because } (-2)^x \text{ is not defined for non-integer } x. \]

3. As \( x \) approaches 4 from the right (i.e., \( x \to 4^+ \)), \( x - 6 \) approaches -2, and since \( x \) is slightly greater than 4, the exponent \( x \) is still positive. Thus:
\[ L_1 \text{ does not exist as } x \to 4^+ \text{ because } (-2)^x \text{ is not defined for non-integer } x. \]

### Conclusion for \( L_1 \):
Since \( L_1 \) does not exist from either side, we conclude:
\[ L_1 \text{ does not exist.} \]

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

1. Substitute \( x = 4 \):
\[ L_2 = (4 - 6)^4 = (-2)^4 = 16 \]
Here, since the exponent is a positive integer, the limit exists.

### Conclusion for \( L_2 \):
\[ L_2 \text{ exists and is equal to } 16. \]

### Final Result:
- \( L_1 \) does not exist.
- \( L_2 \) exists.

### Answer:
The correct option is that \( L_2 \) exists but \( L_1 \) does not exist.