The value of `lambda` that makes the function `f(x) = {{:((cos x)^(1//sinx), x ne 0),(lambda, x =0):}` continuous at x = 0 is:

Options

A. 0
B. 1
C. `1//2`
D. none of these

Answer

To find the value of \(\lambda\) that makes the function

\[ f(x) = \begin{cases} \left(\cos x\right)^{\frac{1}{\sin x}}, & x \neq 0 \\ \lambda, & x = 0 \end{cases} \]

continuous at \(x = 0\), we need to ensure that:

\[ f(0) = \lim_{x \to 0} f(x) \]

This means we need to find the limit of \(f(x)\) as \(x\) approaches \(0\) and set it equal to \(\lambda\).

### Step 1: Calculate the limit as \(x\) approaches \(0\)

We need to evaluate:

\[ \lim_{x \to 0} \left(\cos x\right)^{\frac{1}{\sin x}} \]

### Step 2: Rewrite the limit

This limit is of the form \(1^\infty\) as \(x\) approaches \(0\). To resolve this, we can take the natural logarithm:

Let

\[ L = \lim_{x \to 0} \left(\cos x\right)^{\frac{1}{\sin x}} \]

Taking the natural logarithm:

\[ \ln L = \lim_{x \to 0} \frac{\ln(\cos x)}{\sin x} \]

### Step 3: Apply L'Hôpital's Rule

Since both the numerator and denominator approach \(0\) as \(x \to 0\), we can use L'Hôpital's Rule:

\[ \ln L = \lim_{x \to 0} \frac{\frac{d}{dx}(\ln(\cos x))}{\frac{d}{dx}(\sin x)} \]

Calculating the derivatives:

- The derivative of \(\ln(\cos x)\) is \(-\tan x\).
- The derivative of \(\sin x\) is \(\cos x\).

Thus, we have:

\[ \ln L = \lim_{x \to 0} \frac{-\tan x}{\cos x} = \lim_{x \to 0} -\frac{\sin x}{\cos^2 x} \]

### Step 4: Evaluate the limit

As \(x \to 0\):

\[ \ln L = -\frac{0}{1} = 0 \]

Thus,

\[ L = e^0 = 1 \]

### Step 5: Set \(\lambda\) equal to the limit

Since we want \(f(0) = \lim_{x \to 0} f(x)\), we have:

\[ \lambda = 1 \]

### Conclusion

The value of \(\lambda\) that makes the function continuous at \(x = 0\) is:

\[ \lambda = 1 \]