If the function `f(x) = {((sin^(2)ax)/(x^(2))","," when "x != 0),(" k,"," when " x = 0):}`is continuous at x = 0 then k = ?

Options

A. a
B. `a^(2)`
C. `-2`
D. `-4`

Explanation

To find the value of \( k \) such that the function

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

is continuous at \( x = 0 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is equal to \( f(0) \).

### Step 1: Find the limit as \( x \) approaches 0

We need to calculate:

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

### Step 2: Use the limit property of sine

We can use the known limit:

\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]

To rewrite our limit, we can express \( \sin(ax) \) in terms of \( ax \):

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

### Step 3: Evaluate the limit

As \( x \) approaches 0, \( \frac{\sin(ax)}{ax} \) approaches 1. Therefore, we have:

\[ \lim_{x \to 0} \frac{\sin^2(ax)}{x^2} = 1^2 \cdot a^2 = a^2 \]

### Step 4: Set the limit equal to \( k \)

For the function to be continuous at \( x = 0 \), we need:

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

This gives us:

\[ a^2 = k \]

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{a^2} \]