Sum Rule

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:

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