`lim_(x rarr 0) (cosec x )^(1//"log"x)` is equal to

Options

A. 0
B. 1
C. 1/e
D. none of these

Answer

To solve the limit \( \lim_{x \to 0} (\csc x)^{\frac{1}{\log x}} \), we can follow these steps:

### Step 1: Rewrite the limit
Let \( y = \lim_{x \to 0} (\csc x)^{\frac{1}{\log x}} \).

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we have:
\[ \log y = \lim_{x \to 0} \frac{\log(\csc x)}{\log x} \]

### Step 3: Simplify the logarithm
Using the identity \( \csc x = \frac{1}{\sin x} \), we can rewrite:
\[ \log(\csc x) = \log\left(\frac{1}{\sin x}\right) = -\log(\sin x) \]
Thus, we have:
\[ \log y = \lim_{x \to 0} \frac{-\log(\sin x)}{\log x} \]

### Step 4: Evaluate the limit
As \( x \to 0 \), both \( \log(\sin x) \) and \( \log x \) approach \(-\infty\), resulting in an indeterminate form \(\frac{\infty}{\infty}\). We can apply L'Hôpital's rule:
\[ \log y = \lim_{x \to 0} \frac{-\log(\sin x)}{\log x} \]

### Step 5: Apply L'Hôpital's Rule
Differentiating the numerator and denominator:
- The derivative of \(-\log(\sin x)\) is \(-\frac{\cos x}{\sin x} = -\cot x\).
- The derivative of \(\log x\) is \(\frac{1}{x}\).

Thus, we have:
\[ \log y = \lim_{x \to 0} \frac{-\cot x}{\frac{1}{x}} = \lim_{x \to 0} -x \cot x \]

### Step 6: Simplify the limit
We know that \(\cot x = \frac{\cos x}{\sin x}\), so:
\[ -x \cot x = -x \cdot \frac{\cos x}{\sin x} = -\frac{x \cos x}{\sin x} \]
As \( x \to 0 \), \(\frac{x}{\sin x} \to 1\) and \(\cos x \to 1\). Therefore:
\[ \lim_{x \to 0} -\frac{x \cos x}{\sin x} = -1 \]

### Step 7: Conclude the limit
Thus, we have:
\[ \log y = -1 \implies y = e^{-1} = \frac{1}{e} \]

### Final Answer
\[ \lim_{x \to 0} (\csc x)^{\frac{1}{\log x}} = \frac{1}{e} \]

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