The expression in the image is:

$\lim_{x \to y} \frac{\sin^2 x - \sin^2 y}{x^2 - y^2}$

To solve this limit, we can use some algebraic manipulations and calculus techniques. Let's proceed with these steps:

Step 1: Simplify Using the Difference of Squares

Notice that both the numerator and the denominator can be written as differences of squares:

$\sin^2 x - \sin^2 y = (\sin x + \sin y)(\sin x - \sin y)$ $x^2 - y^2 = (x + y)(x - y)$

Substitute these into the expression:

$\lim_{x \to y} \frac{(\sin x + \sin y)(\sin x - \sin y)}{(x + y)(x - y)}$

Step 2: Cancel Out $(x - y)$

As $x \to y$, both $\sin x - \sin y$ and $x - y$ approach zero. We can apply the limit by dividing these terms. Using the limit property for $\sin x - \sin y$:

$\lim_{x \to y} \frac{\sin x - \sin y}{x - y} = \cos y$

So we can rewrite the limit as:

$\lim_{x \to y} \frac{(\sin x + \sin y) \cdot \cos y}{x + y}$

Step 3: Substitute $x = y$

Now, we can directly substitute $x = y$:

$= \frac{(\sin y + \sin y) \cdot \cos y}{y + y} = \frac{2 \sin y \cdot \cos y}{2y}$

Simplify further:

$= \frac{\sin y \cdot \cos y}{y}$

Final Answer

Thus, the final answer is:

$\frac{\sin y \cdot \cos y}{y}$

Would you like further details or have any questions?

Related Questions

1. How does L'Hôpital's Rule apply to similar limit problems?
2. What are some other techniques to simplify trigonometric limits?
3. How can we evaluate limits involving trigonometric functions in higher dimensions?
4. What happens if we change $\sin^2 x$ to another trigonometric function?
5. Can we solve this limit using Taylor series expansion?

Tip

In limits involving indeterminate forms, always look for algebraic simplifications like factoring or using identities. This can often simplify or even eliminate the need for advanced techniques like L'Hôpital's Rule.