If $$\lim_{x \to 0} \frac{e^{ax} - \cos(bx) - \frac{cxe^{-cx}}{2}}{1 - \cos(2x)} = 17$$, then $$5a^2 + b^2$$ is equal to

We need $$\lim_{x \to 0} \frac{e^{ax} - \cos(bx) - \frac{cx}{2}e^{-cx}}{1 - \cos(2x)} = 17$$

Using Taylor series near $$x = 0$$:

$$e^{ax} \approx 1 + ax + \frac{a^2x^2}{2}$$

$$\cos(bx) \approx 1 - \frac{b^2x^2}{2}$$

$$\frac{cx}{2}e^{-cx} \approx \frac{cx}{2}(1 - cx) = \frac{cx}{2} - \frac{c^2x^2}{2}$$

$$1 - \cos(2x) \approx 2x^2$$

Numerator ≈ $$(1 + ax + \frac{a^2x^2}{2}) - (1 - \frac{b^2x^2}{2}) - (\frac{cx}{2} - \frac{c^2x^2}{2})$$

$$= ax + \frac{a^2x^2}{2} + \frac{b^2x^2}{2} - \frac{cx}{2} + \frac{c^2x^2}{2}$$

For the limit to exist (not $$\pm\infty$$), the coefficient of $$x$$ in the numerator must be 0:

$$a - \frac{c}{2} = 0$$, so $$c = 2a$$

Coefficient of $$x^2$$: $$\frac{a^2 + b^2 + c^2}{2} = \frac{a^2 + b^2 + 4a^2}{2} = \frac{5a^2 + b^2}{2}$$

$$\lim = \frac{(5a^2+b^2)/2}{2} = \frac{5a^2+b^2}{4} = 17$$

$$5a^2 + b^2 = 68$$

This matches option 3: 68.