Let `f(x)={(1-cosx)/(x^2),\ \ \ w h e n\ x!=0 1,\ \ \ \ w h e n\ x=0` . Show that `f(x)` is discontinuous at `x=0` .

Explanation

To show that the function \( f(x) = \frac{1 - \cos x}{x^2} \) when \( x \neq 0 \) and \( f(0) = 1 \) is discontinuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 and compare it to \( f(0) \).

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

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