Value of a so that `f(x) = (sin^(2)ax)/(x^(2)), x != 0 ` and f(0) = 1 is continuous at x = 0 is

Options

A. 0
B. 1
C. `-1`
D. `pm1`

Answer

To determine the value of \( a \) such that the function

\[ f(x) = \frac{\sin^2(ax)}{x^2} \quad \text{for } x \neq 0 \]
and \( f(0) = 1 \) is continuous at \( x = 0 \), we need to ensure that:

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

### Step 1: Set up the limit condition
We start by setting up the limit:

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

### Step 2: Identify the indeterminate form
As \( x \to 0 \), both the numerator and denominator approach 0, leading to the indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
To resolve this, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivative of the numerator and the derivative of the denominator:

1. Derivative of the numerator \( \sin^2(ax) \):
\[ \frac{d}{dx}(\sin^2(ax)) = 2\sin(ax)\cos(ax) \cdot a = a \sin(2ax) \]

2. Derivative of the denominator \( x^2 \):
\[ \frac{d}{dx}(x^2) = 2x \]

Thus, applying L'Hôpital's Rule gives us:

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

### Step 4: Apply L'Hôpital's Rule again (if necessary)
As \( x \to 0 \), we again have an indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule again:

1. Derivative of the numerator \( a \sin(2ax) \):
\[ \frac{d}{dx}(a \sin(2ax)) = 2a^2 \cos(2ax) \]

2. Derivative of the denominator \( 2x \):
\[ \frac{d}{dx}(2x) = 2 \]

Thus, we have:

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

### Step 5: Set the limit equal to \( f(0) \)
Since we want this limit to equal \( f(0) = 1 \):

\[ a^2 = 1 \]

### Step 6: Solve for \( a \)
Taking the square root of both sides gives us:

\[ a = \pm 1 \]

### Final Answer
Thus, the values of \( a \) that make \( f(x) \) continuous at \( x = 0 \) are:

\[ a = 1 \quad \text{or} \quad a = -1 \]