Important Indeterminate Forms

$$\ln(1 + x^2) = x^2 - \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3} + \dots = x^2 - \frac{x^4}{2} + \frac{x^6}{3} + O(x^8)$$

$$\sin(x) = x - \frac{x^3}{6} + O(x^5)$$

$$\sin^2(x) = \left(x - \frac{x^3}{6}\right)^2 = x^2 - 2(x)\left(\frac{x^3}{6}\right) + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$$

$$\text{Denominator} = \left( x^2 - \frac{x^4}{2} \right) - \left( x^2 - \frac{x^4}{3} \right) + O(x^6)$$

$$\text{Denominator} = \left(-\frac{1}{2} + \frac{1}{3}\right)x^4 + O(x^6) = -\frac{1}{6}x^4 + O(x^6)$$

Since the denominator's leading term is of order $$x^4$$, we must expand the numerator up to the $$x^4$$ term as well.

The numerator has two parts: $$\cos(x)e^{x^2/2}$$ and $$\sqrt{1 - x^4}$$.

• Part A: $$\cos(x) \cdot e^{x^2/2}$$

$$\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$

$$e^{x^2/2} = 1 + \left(\frac{x^2}{2}\right) + \frac{1}{2}\left(\frac{x^2}{2}\right)^2 + O(x^6) = 1 + \frac{x^2}{2} + \frac{x^4}{8} + O(x^6)$$

$$\cos(x) \cdot e^{x^2/2} = \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)\left(1 + \frac{x^2}{2} + \frac{x^4}{8}\right)$$

$$= 1 + \frac{x^2}{2} + \frac{x^4}{8} - \frac{x^2}{2} - \frac{x^4}{4} + \frac{x^4}{24} + O(x^6)$$

$$\left(\frac{1}{8} - \frac{1}{4} + \frac{1}{24}\right)x^4 = \left(\frac{3}{24} - \frac{6}{24} + \frac{1}{24}\right)x^4 = -\frac{2}{24}x^4 = -\frac{1}{12}x^4$$

So, $$\cos(x)e^{x^2/2} = 1 - \frac{1}{12}x^4 + O(x^6)$$.

$$\sqrt{1 - x^4} = 1 - \frac{1}{2}x^4 + O(x^8)$$

$$\text{Numerator} = \left(1 - \frac{1}{12}x^4\right) - \left(1 - \frac{1}{2}x^4\right) + O(x^6)$$

$$\text{Numerator} = \left(-\frac{1}{12} + \frac{1}{2}\right)x^4 + O(x^6) = \left(-\frac{1}{12} + \frac{6}{12}\right)x^4 + O(x^6) = \frac{5}{12}x^4 + O(x^6)$$

$$\lim_{x \to 0} \frac{\frac{5}{12}x^4 + O(x^6)}{-\frac{1}{6}x^4 + O(x^6)}$$

$$\lim_{x \to 0} \frac{\frac{5}{12} + O(x^2)}{-\frac{1}{6} + O(x^2)} = \frac{\frac{5}{12}}{-\frac{1}{6}}$$

$$\frac{5}{12} \cdot (-6) = -\frac{30}{12} = -\frac{5}{2}$$

The limit is of the form $$1^\infty$$ as $$x \to 0$$.

1. Apply the standard formula:
$$L = e^{\lim_{x \to 0} \frac{1}{x^2} \left(2 - \cos x \sqrt{\cos 2x} - 1\right)} = e^{\lim_{x \to 0} \frac{1 - \cos x \sqrt{\cos 2x}}{x^2}}$$

2. Use Taylor series expansion for the trigonometric functions:
$$\cos x = 1 - \frac{x^2}{2} + \dots$$
$$\cos 2x = 1 - \frac{(2x)^2}{2} = 1 - 2x^2 + \dots$$
Using the binomial expansion for the square root:
$$\sqrt{\cos 2x} = (1 - 2x^2)^{1/2} = 1 - \frac{1}{2}(2x^2) = 1 - x^2 + \dots$$

3. Multiply the expansions in the numerator:
$$\cos x \sqrt{\cos 2x} \approx \left(1 - \frac{x^2}{2}\right)(1 - x^2) \approx 1 - x^2 - \frac{x^2}{2} = 1 - \frac{3x^2}{2}$$

4. Evaluate the limit inside the exponent:
$$1 - \cos x \sqrt{\cos 2x} \approx 1 - \left(1 - \frac{3x^2}{2}\right) = \frac{3x^2}{2}$$
Substituting this back gives:
$$L = e^{\lim_{x \to 0} \frac{3x^2 / 2}{x^2}} = e^{3/2}$$