If the function `f(x) = {{:((cos x)^(1//x), x ne 0),(=k, x =0):}`, is continous at x=0, then the value of k is:

Options

A. 1
B. `-1`
C. 0
D. e

Answer

To determine the value of \( k \) such that the function

\[ f(x) = \begin{cases} \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

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

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

### Step 1: Find the limit as \( x \) approaches 0.
We need to evaluate the limit:

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

### Step 2: Analyze the behavior of \( \cos\left(\frac{1}{x}\right) \).
As \( x \) approaches 0, \( \frac{1}{x} \) approaches infinity. The cosine function oscillates between -1 and 1 for all real numbers, meaning:

\[ \cos\left(\frac{1}{x}\right) \text{ does not approach a single value as } x \to 0. \]

### Step 3: Use the Squeeze Theorem.
Since \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can apply the Squeeze Theorem:

\[ -1 \leq \cos\left(\frac{1}{x}\right) \leq 1. \]

Thus, we can conclude:

\[ \lim_{x \to 0} \cos\left(\frac{1}{x}\right) \text{ does not exist in the traditional sense, but it is bounded.} \]

### Step 4: Set the limit equal to \( k \).
For the function to be continuous at \( x = 0 \), we need:

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

Given that \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1, the only way for the function to be continuous at \( x = 0 \) is if \( k \) is set to a value that reflects this oscillation. The most reasonable value is:

\[ k = 1. \]

### Step 5: Conclusion.
Thus, the value of \( k \) that makes the function continuous at \( x = 0 \) is:

\[ \boxed{1}. \]