Important Equivalent Infinitesimals

Let

$$L= \lim_{x\to0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}-x}{x^3}$$

Using binomial expansion,

$$\sqrt{1+t} = 1+\frac t2-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)$$

and

$$\sqrt{1-t} = 1-\frac t2-\frac{t^2}{8}-\frac{t^3}{16}+o(t^3)$$

Subtracting,

$$\sqrt{1+t}-\sqrt{1-t} = t+\frac{t^3}{8}+o(t^3)$$

Put

$$t=\sin x$$

Hence,

$$\sqrt{1+\sin x}-\sqrt{1-\sin x} = \sin x+\frac{\sin^3x}{8}+o(x^3)$$

Using

$$\sin x=x-\frac{x^3}{6}+o(x^3),$$

we get

$$\sin^3x=x^3+o(x^3)$$

Therefore,

$$\sqrt{1+\sin x}-\sqrt{1-\sin x}-x$$

$$= \left(x-\frac{x^3}{6}\right) +\frac{x^3}{8} -x+o(x^3)$$

$$= x^3\left(-\frac16+\frac18\right)+o(x^3)$$

$$= -\frac{x^3}{24}+o(x^3)$$

Hence,

$$L=-\frac1{24}$$

Therefore,

$$48\lambda = 48\left(-\frac1{24}\right)$$

$$=-2$$

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

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

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

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

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

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

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

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

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

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

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

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

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:

$$\lim_{x \to 0} \frac{6 \cdot \left(\frac{27x^4}{2}\right)}{x^4}$$

$$\lim_{x \to 0} \frac{3 \cdot 27x^4}{x^4} = 81$$