`f(x) = {{:(|x|sin\ (1)/(x-a),if x ne 0),(0, if x =a):}` at `x = a`

Explanation

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

1. \( f(a) \) is defined.
2. The left-hand limit \( f(a^-) \) exists.
3. The right-hand limit \( f(a^+) \) exists.
4. \( f(a^-) = f(a) = f(a^+) \).

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

### Step 1: Evaluate \( f(a) \)
Since \( f(x) = 0 \) when \( x = a \):
\[ f(a) = 0 \]

### Step 2: Evaluate the left-hand limit \( f(a^-) \)
We need to find:
\[ f(a^-) = \lim_{x \to a^-} f(x) = \lim_{x \to a^-} |x| \sin\left(\frac{1}{x-a}\right) \]
As \( x \) approaches \( a \) from the left, \( |x| \) approaches \( |a| \) and \( \sin\left(\frac{1}{x-a}\right) \) oscillates between -1 and 1. Therefore, we can express this limit as:
\[ \lim_{x \to a^-} |x| \sin\left(\frac{1}{x-a}\right) = |a| \cdot \lim_{x \to a^-} \sin\left(\frac{1}{x-a}\right) \]
Since \( \sin\left(\frac{1}{x-a}\right) \) oscillates between -1 and 1, we have:
\[ \lim_{x \to a^-} |x| \sin\left(\frac{1}{x-a}\right) = 0 \cdot \text{(bounded value)} = 0 \]
Thus:
\[ f(a^-) = 0 \]

### Step 3: Evaluate the right-hand limit \( f(a^+) \)
Now we find:
\[ f(a^+) = \lim_{x \to a^+} f(x) = \lim_{x \to a^+} |x| \sin\left(\frac{1}{x-a}\right) \]
Similar to the left-hand limit:
\[ \lim_{x \to a^+} |x| \sin\left(\frac{1}{x-a}\right) = |a| \cdot \lim_{x \to a^+} \sin\left(\frac{1}{x-a}\right) \]
Again, since \( \sin\left(\frac{1}{x-a}\right) \) oscillates between -1 and 1:
\[ \lim_{x \to a^+} |x| \sin\left(\frac{1}{x-a}\right) = 0 \cdot \text{(bounded value)} = 0 \]
Thus:
\[ f(a^+) = 0 \]

### Step 4: Check continuity
Now we check if:
\[ f(a^-) = f(a) = f(a^+) = 0 \]
Since all three values are equal to 0, we conclude that the function \( f(x) \) is continuous at \( x = a \).

### Final Conclusion
The function \( f(x) \) is continuous at \( x = a \).

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