To evaluate this limit , we use the Maclaurin series expansion for both trigonometric functions :
1. Expand the outer cosine function with a sine argument:
The standard series expansion for $$\cos(\theta)$$ is:
$$\cos(\theta) = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots$$
Substituting $$\theta = \sin(3x)$$ into the expansion gives:
$$\cos(\sin(3x)) = 1 - \frac{\sin^2(3x)}{2} + \frac{\sin^4(3x)}{24} - \dots$$
2. Substitute the series expansion of sine:
The standard series expansion for $$\sin(3x)$$ is:
$$\sin(3x) = 3x - \frac{(3x)^3}{3!} + \dots = 3x - \frac{9x^3}{2} + \dots$$
Now, calculate the powers of $$\sin(3x)$$ up to the fourth degree:
$$\sin^2(3x) = \left(3x - \frac{9x^3}{2}\right)^2 = 9x^2 - 27x^4 + \dots$$
$$\sin^4(3x) = (3x)^4 = 81x^4 + \dots$$
Substitute these back into the expression for $$\cos(\sin(3x))$$:
$$\cos(\sin(3x)) = 1 - \frac{9x^2 - 27x^4}{2} + \frac{81x^4}{24} - \dots$$
$$\cos(\sin(3x)) = 1 - \frac{9x^2}{2} + \frac{27x^4}{2} + \frac{27x^4}{8} - \dots$$
$$\cos(\sin(3x)) = 1 - \frac{9x^2}{2} + \frac{135x^4}{8} - \dots$$
3. Expand the standard cosine term $$\cos(3x)$$:
Using the standard expansion formula for $$\cos(3x)$$:
$$\cos(3x) = 1 - \frac{(3x)^2}{2} + \frac{(3x)^4}{24} - \dots$$
$$\cos(3x) = 1 - \frac{9x^2}{2} + \frac{81x^4}{24} - \dots$$
$$\cos(3x) = 1 - \frac{9x^2}{2} + \frac{27x^4}{8} - \dots$$
4. Evaluate the difference in the numerator:
Subtracting the two series expressions isolates the leading fourth-degree term:
$$\cos(\sin(3x)) - \cos(3x) = \left(1 - \frac{9x^2}{2} + \frac{135x^4}{8}\right) - \left(1 - \frac{9x^2}{2} + \frac{27x^4}{8}\right)$$
$$\cos(\sin(3x)) - \cos(3x) = \frac{135x^4}{8} - \frac{27x^4}{8} = \frac{108x^4}{8} = \frac{27x^4}{2}$$
5. Compute the final limit value:
Now, substitute this simplified numerator into the original limit expression:
$$\lim_{x \to 0} \frac{6 \cdot \left(\frac{27x^4}{2}\right)}{x^4}$$
$$\lim_{x \to 0} \frac{3 \cdot 27x^4}{x^4} = 81$$
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