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

Explanation

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

1. The left-hand limit as \( x \) approaches \( a \) exists.
2. The right-hand limit as \( x \) approaches \( a \) exists.
3. The value of the function at \( x = a \) is equal to both limits.

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

### Step 1: Calculate the Left-Hand Limit
We need to find the left-hand limit as \( x \) approaches \( a \):
\[ \lim_{x \to a^-} f(x) = \lim_{h \to 0} f(a - h) = \lim_{h \to 0} |a - h| \sin\left(\frac{1}{a - h - a}\right) = \lim_{h \to 0} |a - h| \sin\left(\frac{1}{-h}\right) \]
Since \( a - h \) approaches \( a \) as \( h \) approaches \( 0 \), we can simplify:
\[ = \lim_{h \to 0} (a - h) \sin\left(-\frac{1}{h}\right) = \lim_{h \to 0} (a - h)(-\sin\left(\frac{1}{h}\right)) \]
The sine function oscillates between -1 and 1, so we can write:
\[ = \lim_{h \to 0} (a - h) \cdot (-\sin\left(\frac{1}{h}\right)) \]
As \( h \to 0 \), \( a - h \to a \) and \( -\sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1. Thus, the limit approaches:
\[ = 0 \]

### Step 2: Calculate the Right-Hand Limit
Next, we find the right-hand limit as \( x \) approaches \( a \):
\[ \lim_{x \to a^+} f(x) = \lim_{h \to 0} f(a + h) = \lim_{h \to 0} |a + h| \sin\left(\frac{1}{a + h - a}\right) = \lim_{h \to 0} |a + h| \sin\left(\frac{1}{h}\right) \]
As \( h \to 0 \), \( a + h \) approaches \( a \):
\[ = \lim_{h \to 0} (a + h) \sin\left(\frac{1}{h}\right) \]
Again, the sine function oscillates between -1 and 1:
\[ = \lim_{h \to 0} (a + h) \cdot \sin\left(\frac{1}{h}\right) \]
As \( h \to 0 \), \( a + h \to a \) and \( \sin\left(\frac{1}{h}\right) \) oscillates between -1 and 1, leading to:
\[ = 0 \]

### Step 3: Check the Value of the Function at \( x = a \)
Now we check the value of the function at \( x = a \):
\[ f(a) = 0 \]

### Conclusion
Since both the left-hand limit and the right-hand limit as \( x \) approaches \( a \) are equal to \( 0 \), and \( f(a) = 0 \), we have:
\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) = 0 \]
Thus, the function \( f(x) \) is continuous at \( x = a \).