`lim_(xrarr1^(+))(log_(sqrt2)sqrt2x)^(1/({x}))` where [ ] represents greatest integer function is

Options

A. `sqrt2`
B. e
C. `e^(log_sqrt2 (e ))`
D. none of these

Answer

To solve the limit \( \lim_{x \to 1^{+}} \left( \log_{\sqrt{2}} \sqrt{2x} \right)^{\frac{1}{\{x\}}} \), where \(\{x\}\) represents the fractional part of \(x\), we can follow these steps:

### Step 1: Rewrite the logarithm
We start by rewriting the logarithm in terms of natural logarithm:
\[ \log_{\sqrt{2}} \sqrt{2x} = \frac{\ln(\sqrt{2x})}{\ln(\sqrt{2})} = \frac{\ln(\sqrt{2}) + \ln(x)}{\frac{1}{2} \ln(2)} = \frac{2(\ln(\sqrt{2}) + \ln(x))}{\ln(2)} \]

### Step 2: Simplify the expression
Now, we can simplify the expression:
\[ \left( \log_{\sqrt{2}} \sqrt{2x} \right)^{\frac{1}{\{x\}}} = \left( \frac{2(\ln(\sqrt{2}) + \ln(x))}{\ln(2)} \right)^{\frac{1}{\{x\}}} \]

### Step 3: Analyze the limit as \(x \to 1^{+}\)
As \(x\) approaches \(1\) from the right, the fractional part \(\{x\} = x - 1\) approaches \(0\). Therefore, we need to analyze the limit:
\[ \lim_{x \to 1^{+}} \left( \frac{2(\ln(\sqrt{2}) + \ln(x))}{\ln(2)} \right)^{\frac{1}{x-1}} \]

### Step 4: Apply the exponential limit
This limit is of the form \(1^{\infty}\). We can rewrite it using the exponential function:
\[ = e^{\lim_{x \to 1^{+}} \frac{\ln\left( \frac{2(\ln(\sqrt{2}) + \ln(x))}{\ln(2)} \right)}{x-1}} \]

### Step 5: Evaluate the logarithm
Now we evaluate the logarithm:
\[ \ln\left( \frac{2(\ln(\sqrt{2}) + \ln(x))}{\ln(2)} \right) = \ln(2) + \ln(\ln(\sqrt{2}) + \ln(x)) - \ln(\ln(2)) \]

### Step 6: Differentiate using L'Hôpital's Rule
Since both the numerator and denominator approach \(0\) as \(x \to 1^{+}\), we can apply L'Hôpital's Rule:
\[ \lim_{x \to 1^{+}} \frac{\ln(\ln(\sqrt{2}) + \ln(x))}{x-1} \]

### Step 7: Find the derivative
Taking the derivative of the numerator and denominator:
- The derivative of the numerator is:
\[ \frac{1}{\ln(\sqrt{2}) + \ln(x)} \cdot \frac{1}{x} \]
- The derivative of the denominator is \(1\).

### Step 8: Evaluate the limit
Now we evaluate:
\[ \lim_{x \to 1^{+}} \frac{1}{\ln(\sqrt{2}) + \ln(x)} \cdot \frac{1}{x} \]
At \(x = 1\), this becomes:
\[ \frac{1}{\ln(\sqrt{2}) + 0} = \frac{1}{\ln(\sqrt{2})} \]

### Step 9: Final limit
Thus, the limit evaluates to:
\[ e^{\frac{1}{\ln(\sqrt{2})}} = e^{\frac{2}{\ln(2)}} \]

### Step 10: Apply the greatest integer function
Finally, we need to apply the greatest integer function:
\[ \lfloor e^{\frac{2}{\ln(2)}} \rfloor \]

### Conclusion
After calculating \(e^{\frac{2}{\ln(2)}}\), we find the final answer.