$$\lim_{x \to 0} \left(\left(\dfrac{1-\cos^2(3x)}{\cos^3(4x)}\right)\left(\dfrac{\sin^3(4x)}{(\log_e(2x+1))^5}\right)\right)$$ is equal to

To solve this limit quickly, we use the standard limits as $$x \to 0$$:

• $$\sin(kx) \approx kx$$
• $$\ln(1 + kx) \approx kx$$
• $$\cos(kx) \approx 1$$

The expression is:

$$L = \lim_{x \to 0} \left( \frac{1 - \cos^2(3x)}{\cos^3(4x)} \right) \left( \frac{\sin^3(4x)}{(\ln(2x + 1))^5} \right)$$

1. • $$1 - \cos^2(3x) = \sin^2(3x) \approx (3x)^2 = \mathbf{9x^2}$$
   • $$\cos^3(4x) \approx 1^3 = \mathbf{1}$$
   • $$\sin^3(4x) \approx (4x)^3 = \mathbf{64x^3}$$
   • $$(\ln(2x + 1))^5 \approx (2x)^5 = \mathbf{32x^5}$$
2. $$L = \lim_{x \to 0} \left( \frac{9x^2}{1} \right) \left( \frac{64x^3}{32x^5} \right)$$
3. $$L = 9x^2 \cdot \frac{64x^3}{32x^5} = 9 \cdot \frac{64}{32} \cdot \frac{x^5}{x^5}$$
4. $$L = 9 \cdot 2 = \mathbf{18}$$