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

Explanation

To determine whether the function \( f(x) \) is continuous at \( x = 0 \), we need to check the following:

1. **Value of the function at \( x = 0 \)**:
\[ f(0) = 0 \]

2. **Left-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} |x| \cos\left(\frac{1}{x}\right) \]
Since \( x \) is approaching 0 from the left, we have \( |x| = -x \). Thus, we can rewrite the limit as:
\[ \lim_{h \to 0^+} h \cos\left(\frac{1}{-h}\right) = \lim_{h \to 0^+} h \cos\left(-\frac{1}{h}\right) = \lim_{h \to 0^+} h \cos\left(\frac{1}{h}\right) \]

3. **Evaluating the limit**:
The cosine function oscillates between -1 and 1, so we can bound the limit:
\[ -h \leq h \cos\left(\frac{1}{h}\right) \leq h \]
As \( h \to 0^+ \), both bounds approach 0:
\[ \lim_{h \to 0^+} -h = 0 \quad \text{and} \quad \lim_{h \to 0^+} h = 0 \]
By the Squeeze Theorem:
\[ \lim_{h \to 0^+} h \cos\left(\frac{1}{h}\right) = 0 \]

4. **Conclusion about the left-hand limit**:
Therefore, we have:
\[ \lim_{x \to 0^-} f(x) = 0 \]

5. **Right-hand limit as \( x \) approaches 0**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} |x| \cos\left(\frac{1}{x}\right) = \lim_{h \to 0^+} h \cos\left(\frac{1}{h}\right) \]
Following the same reasoning as above, we find:
\[ \lim_{h \to 0^+} h \cos\left(\frac{1}{h}\right) = 0 \]

6. **Conclusion about the right-hand limit**:
Therefore, we have:
\[ \lim_{x \to 0^+} f(x) = 0 \]

7. **Final check for continuity**:
Since both the left-hand limit and the right-hand limit at \( x = 0 \) are equal to the value of the function at that point:
\[ \lim_{x \to 0} f(x) = f(0) = 0 \]
We conclude that \( f(x) \) is continuous at \( x = 0 \).

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