The expression from the image involves finding the limit:

$\lim_{x \to 0} \frac{\sin(5x) - \sin(3x)}{\sin(x)}$

Let's solve this limit step by step.

Step 1: Simplify the expression

We can use the trigonometric identity for the difference of sines:

$\sin(A) - \sin(B) = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$

Applying this identity to $\sin(5x) - \sin(3x)$:

$\sin(5x) - \sin(3x) = 2 \cos\left(\frac{5x + 3x}{2}\right) \sin\left(\frac{5x - 3x}{2}\right)$

Simplifying further:

$\sin(5x) - \sin(3x) = 2 \cos(4x) \sin(x)$

Step 2: Substitute and simplify the limit expression

Now substitute this into the original limit:

$\lim_{x \to 0} \frac{2 \cos(4x) \sin(x)}{\sin(x)}$

We can cancel out $\sin(x)$ from the numerator and the denominator:

$\lim_{x \to 0} 2 \cos(4x)$

Step 3: Evaluate the limit

As $x$ approaches 0, $\cos(4x)$ approaches $\cos(0) = 1$. Thus, the limit is:

$\lim_{x \to 0} 2 \cos(4x) = 2 \times 1 = 2$

Final Answer:

$\lim_{x \to 0} \frac{\sin(5x) - \sin(3x)}{\sin(x)} = 2$

Would you like to dive into any details or have any questions?

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Here are some related questions:

1. What is the general method for solving trigonometric limits?
2. How can L'Hôpital's rule be applied to limits involving sine functions?
3. Can you derive the identity $\sin(A) - \sin(B) = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$?
4. What are some other common trigonometric identities used in limit problems?
5. How would the limit change if the denominator were $\cos(x)$ instead of $\sin(x)$?
6. What happens if we evaluate the limit at a point other than zero?
7. How can this limit be approached using series expansions of sine and cosine?
8. What are the implications of the small-angle approximation for sine in limit problems?

Tip: Always remember to check for possible simplifications using trigonometric identities before applying more complex limit-solving techniques.