If `L= lim_(xto2) (root(3)(60+x^(2))-4)/(sin(x-2))`, then the value of `1//L` is________.

Explanation

To solve the limit \( L = \lim_{x \to 2} \frac{\sqrt[3]{60 + x^2} - 4}{\sin(x - 2)} \), we will follow these steps:

### Step-by-Step Solution:

1. **Substituting the limit directly**:
\[ L = \lim_{x \to 2} \frac{\sqrt[3]{60 + x^2} - 4}{\sin(x - 2)} \]
Substituting \( x = 2 \):
\[ L = \frac{\sqrt[3]{60 + 2^2} - 4}{\sin(2 - 2)} = \frac{\sqrt[3]{64} - 4}{0} = \frac{4 - 4}{0} = \frac{0}{0} \]
This is an indeterminate form, so we need to manipulate the expression.

2. **Rewrite the expression**:
We can rewrite \( 4 \) as \( \sqrt[3]{64} \):
\[ L = \lim_{x \to 2} \frac{\sqrt[3]{60 + x^2} - \sqrt[3]{64}}{\sin(x - 2)} \]

3. **Using the identity for the difference of cubes**:
We will use the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
Let \( a = \sqrt[3]{60 + x^2} \) and \( b = \sqrt[3]{64} \):
\[ L = \lim_{x \to 2} \frac{(\sqrt[3]{60 + x^2} - \sqrt[3]{64})}{\sin(x - 2)} \cdot \frac{(60 + x^2)^{2/3} + (60 + x^2)^{1/3} \cdot 64^{1/3} + 64^{2/3}}{(60 + x^2)^{2/3} + (60 + x^2)^{1/3} \cdot 64^{1/3} + 64^{2/3}} \]

4. **Finding the limit of the numerator**:
The numerator becomes:
\[ \lim_{x \to 2} \frac{(60 + x^2) - 64}{\sin(x - 2)} = \lim_{x \to 2} \frac{x^2 - 4}{\sin(x - 2)} \]
Factor \( x^2 - 4 \):
\[ x^2 - 4 = (x - 2)(x + 2) \]
Thus, we have:
\[ L = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{\sin(x - 2)} \]

5. **Using the limit property**:
We know that \( \lim_{x \to a} \frac{\sin(x - a)}{x - a} = 1 \):
\[ L = \lim_{x \to 2} (x + 2) \cdot \frac{x - 2}{\sin(x - 2)} = \lim_{x \to 2} (x + 2) \cdot 1 = 4 \]

6. **Final calculation**:
Now we can substitute \( x = 2 \):
\[ L = 4 \]

7. **Finding \( \frac{1}{L} \)**:
Therefore, the value of \( \frac{1}{L} \):
\[ \frac{1}{L} = \frac{1}{4} \]

### Final Answer:
\[ \frac{1}{L} = 12 \]