`f(x)={{:(|x|cos'1/x, if x ne 0),(0, if x =0):}` at `x = 0` .

Explanation

To determine the continuity of the function \( f(x) \) at \( x = 0 \), we need to check the following conditions:

1. \( f(0) \) must be defined.
2. The limit of \( f(x) \) as \( x \) approaches \( 0 \) from the right (denoted as \( f(0^+) \)) must exist.
3. The limit of \( f(x) \) as \( x \) approaches \( 0 \) from the left (denoted as \( f(0^-) \)) must exist.
4. The limits from both sides must be equal to \( f(0) \).

Given the function:
\[ f(x) = \begin{cases} |x| \cos\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

### Step 1: Calculate \( f(0) \)
From the definition of the function, we have:
\[ f(0) = 0 \]

### Step 2: Calculate \( f(0^+) \)
To find \( f(0^+) \), we consider the limit as \( x \) approaches \( 0 \) from the right:
\[ f(0^+) = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x| \cos\left(\frac{1}{x}\right) \]
Since \( |x| = x \) for \( x > 0 \), we can rewrite this as:
\[ f(0^+) = \lim_{x \to 0^+} x \cos\left(\frac{1}{x}\right) \]
As \( x \to 0^+ \), \( \cos\left(\frac{1}{x}\right) \) oscillates between -1 and 1. Therefore, we can bound the limit:
\[ -x \leq x \cos\left(\frac{1}{x}\right) \leq x \]
As \( x \to 0^+ \), both \( -x \) and \( x \) approach \( 0 \). By the Squeeze Theorem:
\[ f(0^+) = 0 \]

### Step 3: Calculate \( f(0^-) \)
To find \( f(0^-) \), we consider the limit as \( x \) approaches \( 0 \) from the left:
\[ f(0^-) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x| \cos\left(\frac{1}{x}\right) \]
Since \( |x| = -x \) for \( x < 0 \), we can rewrite this as:
\[ f(0^-) = \lim_{x \to 0^-} -x \cos\left(\frac{1}{x}\right) \]
Again, using the same reasoning as before:
\[ -x \leq -x \cos\left(\frac{1}{x}\right) \leq x \]
As \( x \to 0^- \), both bounds approach \( 0 \). Thus, by the Squeeze Theorem:
\[ f(0^-) = 0 \]

### Step 4: Check continuity
Now we have:
\[ f(0^+) = 0, \quad f(0) = 0, \quad f(0^-) = 0 \]
Since:
\[ f(0^+) = f(0) = f(0^-) \]
We conclude that the function \( f(x) \) is continuous at \( x = 0 \).

### Final Answer:
The function \( f(x) \) is continuous at \( x = 0 \).

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