Disuss the continutiy of `f(x) ={:{((sin^2 x)/(x^2)", "x ne 0),(" "0", " x= 0):},`
at x=0

Answer

To discuss the continuity of the function

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

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

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

### Step 1: Find \( f(0) \)

From the definition of the function, we have:

\[ f(0) = 0 \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \), we have:

\[ f(x) = \frac{\sin^2 x}{x^2} \]

We need to evaluate the limit:

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

### Step 3: Simplify the limit

Using the identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we can rewrite the limit as:

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

### Step 4: Apply the limit

Now, applying the limit:

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

### Step 5: Compare the limit with \( f(0) \)

Now we have:

\[ \lim_{x \to 0} f(x) = 1 \quad \text{and} \quad f(0) = 0 \]

### Step 6: Conclusion

Since

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

we conclude that the function \( f(x) \) is discontinuous at \( x = 0 \).

### Summary

The function \( f(x) \) is discontinuous at \( x = 0 \) because the limit of \( f(x) \) as \( x \) approaches 0 does not equal the value of the function at that point.

---