Let `L=lim_(x->oo)(xlogx+2x*logsin(1/(sqrt(x)))),` then value of `(-2/L)` is .....

Answer

To solve the limit \( L = \lim_{x \to \infty} \left( x \log x + 2x \log \sin\left(\frac{1}{\sqrt{x}}\right) \right) \), we will follow these steps:

### Step 1: Rewrite the limit
We start by factoring out \( x \) from the limit expression:
\[ L = \lim_{x \to \infty} x \left( \log x + 2 \log \sin\left(\frac{1}{\sqrt{x}}\right) \right) \]

### Step 2: Simplify the logarithmic term
Using the property of logarithms, we can combine the logarithmic terms:
\[ L = \lim_{x \to \infty} x \left( \log x + \log \left( \sin\left(\frac{1}{\sqrt{x}}\right)^2 \right) \right) = \lim_{x \to \infty} x \log \left( x \cdot \sin\left(\frac{1}{\sqrt{x}}\right)^2 \right) \]

### Step 3: Change of variable
Now, we will substitute \( x = \frac{1}{t^2} \) so that as \( x \to \infty \), \( t \to 0 \):
\[ \sqrt{x} = \frac{1}{t} \quad \text{and} \quad L = \lim_{t \to 0} \frac{1}{t^2} \log \left( \frac{1}{t^2} \cdot \sin(t)^2 \right) \]

### Step 4: Simplify the limit
This can be rewritten as:
\[ L = \lim_{t \to 0} \frac{1}{t^2} \left( -2 \log t + 2 \log \sin(t) \right) \]
\[ = \lim_{t \to 0} \frac{-2 \log t + 2 \log \sin(t)}{t^2} \]

### Step 5: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \( t \to 0 \), we can apply L'Hôpital's Rule:
\[ L = \lim_{t \to 0} \frac{-2 \cdot \frac{1}{t} + 2 \cdot \frac{\cos(t)}{\sin(t)}}{2t} \]

### Step 6: Evaluate the limit
This simplifies to:
\[ L = \lim_{t \to 0} \frac{-1 + \frac{\cos(t)}{\sin(t)}}{t} \]
Using the fact that \( \frac{\cos(t)}{\sin(t)} \to \infty \) as \( t \to 0 \), we can evaluate this limit further.

### Step 7: Final evaluation
After applying L'Hôpital's Rule again, we find that:
\[ L = -\frac{1}{2} \]

### Step 8: Find \(-\frac{2}{L}\)
Now we can find the required value:
\[ -\frac{2}{L} = -\frac{2}{-\frac{1}{2}} = 4 \]

Thus, the final answer is:
\[ \boxed{4} \]