the value of `lim_(x->y) (sin^2x-sin^2y)/(x^2-y^2)` equals

Answer

To solve the limit \( \lim_{x \to y} \frac{\sin^2 x - \sin^2 y}{x^2 - y^2} \), we can follow these steps:

### Step 1: Recognize the Form
First, we substitute \( x = y \) into the expression. This gives us:
\[ \frac{\sin^2 y - \sin^2 y}{y^2 - y^2} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator separately with respect to \( x \).

**Numerator:**
\[ \frac{d}{dx}(\sin^2 x - \sin^2 y) = 2\sin x \cos x = \sin(2x) \]
(Note: The derivative of \( \sin^2 y \) is 0 since \( y \) is treated as a constant.)

**Denominator:**
\[ \frac{d}{dx}(x^2 - y^2) = 2x \]

### Step 3: Rewrite the Limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to y} \frac{\sin(2x)}{2x} \]

### Step 4: Substitute the Limit
Now we substitute \( x = y \) into the limit:
\[ \frac{\sin(2y)}{2y} \]

### Final Answer
Thus, the value of the limit is:
\[ \lim_{x \to y} \frac{\sin^2 x - \sin^2 y}{x^2 - y^2} = \frac{\sin(2y)}{2y} \]