To solve the limit [tex]\(\lim _{x \rightarrow y} \frac{\sin ^2 x-\sin ^2 y}{x^2-y^2}\)[/tex], we can proceed with the following steps:
1. Recognize the difference of squares:
The numerator [tex]\(\sin^2 x - \sin^2 y\)[/tex] can be factored as:
[tex]\[
\sin^2 x - \sin^2 y = (\sin x - \sin y)(\sin x + \sin y)
\][/tex]
Similarly, the denominator [tex]\(x^2 - y^2\)[/tex] can be factored as:
[tex]\[
x^2 - y^2 = (x - y)(x + y)
\][/tex]
2. Rewrite the fraction:
Using the factorizations from step 1, we can rewrite the limit expression:
[tex]\[
\frac{\sin^2 x - \sin^2 y}{x^2 - y^2} = \frac{(\sin x - \sin y)(\sin x + \sin y)}{(x - y)(x + y)}
\][/tex]
3. Cancel common terms:
As long as [tex]\(x \neq y\)[/tex], we can cancel the [tex]\((x - y)\)[/tex] term from the numerator and denominator:
[tex]\[
\frac{(\sin x - \sin y)(\sin x + \sin y)}{(x - y)(x + y)} = \frac{\sin x - \sin y}{x - y} \cdot \frac{\sin x + \sin y}{x + y}
\][/tex]
4. Use the definition of the derivative:
Notice that the first term [tex]\(\frac{\sin x - \sin y}{x - y}\)[/tex] can be recognized as the difference quotient of [tex]\(\sin x\)[/tex] at [tex]\(x = y\)[/tex]. As [tex]\(x\)[/tex] approaches [tex]\(y\)[/tex], this term approaches the derivative of [tex]\(\sin x\)[/tex] at [tex]\(x = y\)[/tex]:
[tex]\[
\lim_{x \to y} \frac{\sin x - \sin y}{x - y} = \cos y
\][/tex]
The second term [tex]\(\frac{\sin x + \sin y}{x + y}\)[/tex] approaches [tex]\(\frac{2 \sin y}{2y} = \frac{\sin y}{y}\)[/tex] as [tex]\(x\)[/tex] approaches [tex]\(y\)[/tex]:
[tex]\[
\lim_{x \to y} \frac{\sin x + \sin y}{x + y} = \frac{\sin y}{y}
\][/tex]
5. Combine the results:
Now, we can combine these results together:
[tex]\[
\lim_{x \to y} \frac{\sin^2 x - \sin^2 y}{x^2 - y^2} = \left(\cos y\right) \cdot \left(\frac{\sin y}{y}\right)
\][/tex]
Thus, the limit evaluates to:
[tex]\[
\boxed{\frac{\sin(2y)}{2y}}
\][/tex]
