The question provided in the image is:

$\lim_{h \to 0} \frac{\cos(3(x+2h)) - \cos(3x)}{h}.$

Step-by-step solution:

This limit can be solved using trigonometric identities and derivatives.

1. Use the trigonometric identity for the difference of cosines:

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

Here:

• $A = 3(x + 2h)$,
• $B = 3x$.

Substitute these into the identity: $\cos(3(x+2h)) - \cos(3x) = -2 \sin\left(\frac{3(x+2h) + 3x}{2}\right) \sin\left(\frac{3(x+2h) - 3x}{2}\right).$

2. Simplify each term:

• $\frac{3(x+2h) + 3x}{2} = \frac{3x + 6h + 3x}{2} = 3x + 3h$,
• $\frac{3(x+2h) - 3x}{2} = \frac{3x + 6h - 3x}{2} = 3h$.

So: $\cos(3(x+2h)) - \cos(3x) = -2 \sin(3x + 3h) \sin(3h).$

3. Substitute into the limit:

$\lim_{h \to 0} \frac{\cos(3(x+2h)) - \cos(3x)}{h} = \lim_{h \to 0} \frac{-2 \sin(3x + 3h) \sin(3h)}{h}.$

4. Separate the terms:

Rewrite as: $\lim_{h \to 0} \frac{-2 \sin(3x + 3h) \sin(3h)}{h} = -2 \lim_{h \to 0} \left[\sin(3x + 3h) \cdot \frac{\sin(3h)}{h}\right].$

5. Evaluate each part:

• As $h \to 0$, $\sin(3x + 3h) \to \sin(3x)$ because $3h \to 0$.
• $\frac{\sin(3h)}{h} \to 3$, because $\lim_{u \to 0} \frac{\sin(u)}{u} = 1$.

Substitute these limits: $-2 \lim_{h \to 0} \left[\sin(3x + 3h) \cdot \frac{\sin(3h)}{h}\right] = -2 \sin(3x) \cdot 3.$

6. Final answer:

$-2 \sin(3x) \cdot 3 = -6 \sin(3x).$

Final Answer:

$\boxed{-6 \sin(3x)}$

Let me know if you would like further clarification or assistance!

Related Questions to Explore:

1. How does the limit definition relate to the derivative of a function?
2. What is the trigonometric identity for $\cos(A) - \cos(B)$, and how is it derived?
3. How do you evaluate limits involving sine or cosine functions?
4. What is the significance of the derivative of $\cos(kx)$?
5. How can this type of problem be solved using Taylor series approximations?

Tip: Always check for trigonometric identities when simplifying expressions involving trigonometric functions in limits.