How to prove this strange limit?

Note that with $\alpha = e^{i \pi/3}$ and $\beta = e^{-i \pi/3}$ we have $\alpha \beta = 1$ and $\alpha + \beta = 1$ and , therefore,

$$\tag{1}f(x) + f'(x) + f''(x) = \alpha\beta f(x) + ( \alpha + \beta)f'(x) + f''(x) \\ =\alpha[ \, \beta f(x) + f'(x) + (\beta f(x) + f'(x))' \, ]$$

One can prove the lemma (when the real part of $\gamma$ is positive):

$$\gamma f(x) + f'(x) \to \delta \implies f(x) \to \delta/\gamma$$

To prove the lemma use the Hardy - L'Hospital trick

$$\lim_{x \to \infty}f(x) = \lim_{x \to \infty}\frac{e^{\gamma x}f(x)}{e^{\gamma x}} = \lim_{x \to \infty}\frac{e^{\gamma x}(\gamma f(x) + f'(x))}{\gamma e^{\gamma x}} = \frac{\delta}{\gamma}.$$

Note that at this stage to appy L'Hospital's rule we don't need to assume anything about the existence of the limit of $f(x)$ in the numerator, only that the limit of the denominator is $+\infty.$

Now by (1) and the lemma we have

$$f(x) + f'(x) + f''(x) \to a \implies \beta f(x) + f'(x) \to a/\alpha,$$

and using the lemma again,

$$f(x) \to a/(\alpha \beta) = a$$