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

Explanation

To check the continuity of the function

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

at \( x = 0 \), we need to verify the following conditions:

1. The left-hand limit as \( x \) approaches 0.
2. The right-hand limit as \( x \) approaches 0.
3. The value of the function at \( x = 0 \).

A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point are all equal.

### Step 1: Calculate the Left-Hand Limit

The left-hand limit as \( x \) approaches 0 is given by:

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

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

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

Since \( \cos(-\theta) = \cos(\theta) \):

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

Using the trigonometric identity \( 1 - \cos(2h) = 2 \sin^2(h) \):

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

Using the standard limit \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):

\[ = 2 \cdot 1^2 = 2 \]

Thus, the left-hand limit is:

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

### Step 2: Calculate the Right-Hand Limit

The right-hand limit as \( x \) approaches 0 is given by:

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

Similarly, we substitute \( x \) with \( 0 + h \) (where \( h \to 0^+ \)):

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

Using the same trigonometric identity as before:

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

Again, using the standard limit \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):

\[ = 2 \cdot 1^2 = 2 \]

Thus, the right-hand limit is:

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

### Step 3: Value of the Function at \( x = 0 \)

Now, we check the value of the function at \( x = 0 \):

\[ f(0) = 5 \]

### Conclusion

Now we compare the limits and the function value:

- Left-hand limit: \( 2 \)
- Right-hand limit: \( 2 \)
- Function value at \( x = 0 \): \( 5 \)

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

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

Thus, the function \( f(x) \) is **not continuous** at \( x = 0 \).