` f(x)={{:((1-cos2x)/(x^(2)),if x ne 0),(5, if x = 0):}` at `x = 0`.

Explanation

To determine whether the function \( f(x) \) is continuous at \( x = 0 \), we need to check the left-hand limit and see if it equals \( f(0) \).

The function is defined as:
\[ f(x) = \begin{cases} \frac{1 - \cos(2x)}{x^2} & \text{if } x \neq 0 \\ 5 & \text{if } x = 0 \end{cases} \]

### Step 1: Find the left-hand limit as \( x \) approaches 0.
We need to calculate:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos(2x)}{x^2} \]

### Step 2: Substitute \( x \) with \( -h \) (where \( h \to 0^+ \)).
This gives us:
\[ \lim_{h \to 0^+} \frac{1 - \cos(-2h)}{(-h)^2} = \lim_{h \to 0^+} \frac{1 - \cos(2h)}{h^2} \]

### Step 3: Use the identity for \( \cos(2h) \).
Recall that:
\[ 1 - \cos(2h) = 2 \sin^2(h) \]
Thus, we can rewrite the limit:
\[ \lim_{h \to 0^+} \frac{2 \sin^2(h)}{h^2} \]

### Step 4: Simplify the limit.
This can be simplified using the limit property:
\[ \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \]
So we have:
\[ \lim_{h \to 0^+} \frac{2 \sin^2(h)}{h^2} = 2 \cdot \left( \lim_{h \to 0} \frac{\sin(h)}{h} \right)^2 = 2 \cdot 1^2 = 2 \]

### Step 5: Compare the left-hand limit to \( f(0) \).
We found that:
\[ \lim_{x \to 0^-} f(x) = 2 \]
And since \( f(0) = 5 \), we see that:
\[ \lim_{x \to 0^-} f(x) \neq f(0) \]

### Conclusion
Since the left-hand limit does not equal \( f(0) \), the function \( f(x) \) is discontinuous at \( x = 0 \).

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

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