The value of k for which `f(x)={{:(,[1+x(e^(-1//x^(2)))sin(1/(x^(4)))]^(e^(1//x^(2))),x ne 0),(,k,x=0):}` is continuous at x=0, is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} (1 + x e^{-\frac{1}{x^2}})^{e^{\frac{1}{x^2}}}, & x \neq 0 \\ k, & x = 0 \end{cases} \]

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

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

### Step 1: Find \( f(0) \)
From the definition of the function, we have:

\[ f(0) = k. \]

### Step 2: Evaluate the limit \( \lim_{x \to 0} f(x) \)
We need to evaluate:

\[ \lim_{x \to 0} (1 + x e^{-\frac{1}{x^2}})^{e^{\frac{1}{x^2}}}. \]

### Step 3: Analyze the expression
As \( x \to 0 \):

- \( x e^{-\frac{1}{x^2}} \to 0 \) because \( e^{-\frac{1}{x^2}} \) approaches 0 much faster than \( x \) approaches 0.
- Thus, \( 1 + x e^{-\frac{1}{x^2}} \to 1 \).

This means we have an expression of the form \( 1^{\infty} \).

### Step 4: Use the limit transformation
To resolve the \( 1^{\infty} \) form, we can rewrite the limit using the exponential function:

\[ \lim_{x \to 0} (1 + x e^{-\frac{1}{x^2}})^{e^{\frac{1}{x^2}}} = \lim_{x \to 0} e^{e^{\frac{1}{x^2}} \ln(1 + x e^{-\frac{1}{x^2}})}. \]

### Step 5: Simplify \( \ln(1 + x e^{-\frac{1}{x^2}}) \)
Using the approximation \( \ln(1 + u) \approx u \) for small \( u \):

\[ \ln(1 + x e^{-\frac{1}{x^2}}) \approx x e^{-\frac{1}{x^2}}. \]

### Step 6: Substitute back into the limit
Now we have:

\[ \lim_{x \to 0} e^{e^{\frac{1}{x^2}} \cdot x e^{-\frac{1}{x^2}}} = \lim_{x \to 0} e^{x}. \]

### Step 7: Evaluate the limit
As \( x \to 0 \):

\[ e^{x} \to e^{0} = 1. \]

### Step 8: Set the limit equal to \( f(0) \)
Now we have:

\[ \lim_{x \to 0} f(x) = 1. \]

For continuity at \( x = 0 \):

\[ f(0) = k \implies k = 1. \]

### Conclusion
Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{1}. \]