Derivative of Exponential Function

Recall that $$\cosh(x) = \frac{e^x + e^{-x}}{2}$$ and use Taylor series expansions around $$x = 0$$.
Taylor series:

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + O(x^6)$$

$$e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120} + O(x^6)$$

Calculate $$\cosh(x)$$:
$$\cosh(x) = \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$$

Calculate the numerator:
$$\cosh(x) - 1 - \frac{x^2}{2} = 1 + \frac{x^2}{2} + \frac{x^4}{24} - 1 - \frac{x^2}{2} + O(x^6)$$
$$= \frac{x^4}{24} + O(x^6)$$

Evaluate the limit:
$$\lim_{x \to 0} \frac{\frac{x^4}{24} + O(x^6)}{x^4} = \lim_{x \to 0} \left(\frac{1}{24} + O(x^2)\right) = \frac{1}{24}$$

Expand the functions in the numerator using Taylor series around zero:

$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$$

$$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$$

Substitute these series back into the numerator:

$$\text{Numerator} = a\left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) + b\left(1 + x^2 + \frac{x^4}{2}\right) - 3$$

Group the terms by powers of $$x$$:

$$\text{Numerator} = (a + b - 3) + \left(b - \frac{a}{2}\right)x^2 + \left(\frac{a}{24} + \frac{b}{2}\right)x^4 + \dots$$

For the limit to be finite when dividing by $$x^4$$, the coefficients of $$x^0$$ and $$x^2$$ must be exactly zero:

1) $$a + b - 3 = 0 \implies a + b = 3$$

2) $$b - \frac{a}{2} = 0 \implies a = 2b$$

Substitute $$a = 2b$$ into the first condition:

$$2b + b = 3 \implies 3b = 3 \implies b = 1$$.