TITLE: How to prove this strange limit?
QUESTION [6 upvotes]: Let $f:[0,\infty)\to\mathbb R$ be a function in $C^2$ such that 
$\lim_{x\to\infty} (f(x)+f'(x)+f''(x)) = a.$
Prove that $\lim_{x\to\infty} f(x)=a$

REPLY [3 votes]: 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, 
$$f(x) \to a/(\alpha \beta) = a$$