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

Explanation

To determine whether the function

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

is continuous at \( x = 0 \), we need to check the left-hand limit, right-hand limit, and the value of the function at that point.

### Step 1: Calculate the Left-Hand Limit
We start by calculating the left-hand limit as \( x \) approaches 0:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos(2x)}{x^2} \]

To evaluate this limit, we can substitute \( x \) with \( -h \) (where \( h \to 0^+ \)):

\[ \lim_{h \to 0} \frac{1 - \cos(-2h)}{(-h)^2} = \lim_{h \to 0} \frac{1 - \cos(2h)}{h^2} \]

### Step 2: Use the Cosine Identity
Using the identity \( 1 - \cos(2h) = 2\sin^2(h) \):

\[ \lim_{h \to 0} \frac{2\sin^2(h)}{h^2} \]

### Step 3: Simplify the Limit
We can rewrite the limit as:

\[ \lim_{h \to 0} 2 \cdot \frac{\sin^2(h)}{h^2} = 2 \cdot \lim_{h \to 0} \left(\frac{\sin(h)}{h}\right)^2 \]

Since \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):

\[ \lim_{h \to 0} 2 \cdot 1^2 = 2 \]

Thus, the left-hand limit is:

\[ \lim_{x \to 0^-} f(x) = 2 \]

### Step 4: Calculate the Right-Hand Limit
Now, we calculate the right-hand limit as \( x \) approaches 0:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1 - \cos(2x)}{x^2} \]

Using the same substitution \( x = h \) (where \( h \to 0^+ \)):

\[ \lim_{h \to 0} \frac{1 - \cos(2h)}{h^2} \]

Using the same cosine identity:

\[ \lim_{h \to 0} \frac{2\sin^2(h)}{h^2} = 2 \cdot \lim_{h \to 0} \left(\frac{\sin(h)}{h}\right)^2 = 2 \cdot 1^2 = 2 \]

Thus, the right-hand limit is:

\[ \lim_{x \to 0^+} f(x) = 2 \]

### Step 5: Compare Limits and Function Value
Now we compare the limits and the function value at \( x = 0 \):

- Left-hand limit: \( 2 \)
- Right-hand limit: \( 2 \)
- Value of the function at \( x = 0 \): \( f(0) = 5 \)

### Conclusion
Since the left-hand limit and right-hand limit are equal but do not equal the value of the function at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = 2 \neq f(0) = 5 \]

Thus, the function \( f(x) \) is **discontinuous** at \( x = 0 \).

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