Discuss the continuity of the function :
`f(x)={{:((1-cosx)/x^(2)", "xne0),(1", "x=0):}` at x = 0.

Answer

To determine the continuity of the function

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

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

1. **Evaluate \( f(0) \)**:
\[ f(0) = 1 \]

2. **Find the limit of \( f(x) \) as \( x \) approaches 0**:
We need to compute:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \]

To evaluate this limit, we can use the trigonometric identity:
\[ 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2} \]

We can also express \( x^2 \) in terms of \( \left(\frac{x}{2}\right)^2 \):
\[ x^2 = 4 \left(\frac{x}{2}\right)^2 \]
Therefore, the limit becomes:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{4 \left(\frac{x}{2}\right)^2} = \lim_{x \to 0} \frac{1}{2} \cdot \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \]

We know that:
\[ \lim_{u \to 0} \frac{\sin u}{u} = 1 \]
where \( u = \frac{x}{2} \). Thus:
\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} = 1 \]

Therefore, we have:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

3. **Compare the limit with \( f(0) \)**:
We found that:
\[ \lim_{x \to 0} f(x) = \frac{1}{2} \]
and
\[ f(0) = 1 \]

Since the limit as \( x \) approaches 0 is not equal to \( f(0) \):
\[ \lim_{x \to 0} f(x) \neq f(0) \]

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