Result is valid if we change $0$ by $a \in \mathbb{R}^*$?

Suppose that $f$ is 2-differentiable and that $\lim_{x \rightarrow \infty} (f(x)+f'(x)+f''(x))=0$. Show that $\lim_{x \rightarrow \infty} f(x) =0$. My question is: result is valid if we change $0$ by $a \in \mathbb{R}^*$?

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Excellent! @Davide Giraudo. – Mario De León Urbina Oct 9 '11 at 18:58
Perhaps, Davide, you would like to post your comment as an answer? – Gerry Myerson Oct 9 '11 at 21:45

The most difficult is to show the result, since for the generalization, if we take $f\in \mathcal C^2(\mathbb R)$ such that $\displaystyle\lim_{x\to +\infty}f(x)+f'(x)+f''(x)=a$ then considering $g(x):=f(x)-a$ we get that $\displaystyle\lim_{x\to +\infty}g(x)+g'(x)+g''(x)=0$ since $g'(x)=f'(x)$ and $g''(x)=f''(x)$, hence applying the previous case we have $\displaystyle\lim_{x\to +\infty}g(x)=0$, therefore $\displaystyle\lim_{x\to +\infty}f(x)=a$.
To show the result, we can write $f(x)+f'(x)+f''(x)=\varepsilon(x)$ where $\varepsilon$ is a continuous function such that $\displaystyle\lim_{x\to+\infty}\varepsilon(x)=0$. We will solve this differential equation. Note that the solutions of the homogenous equation are given by $Ae^{jx}+Be^{\bar j x}$ where $j$ is a complex number such that $j^2+j+1=0$ and $A$ and $B$ are constants. Now, we will try to find a particular solution of the equation. We will try the form $u(x)=A(x)e^{jx}$. Then we should have $u''(x)+u'(x)+u(x)=\varepsilon(x)$, hence $$A''(x)e^{jx}+2A'(x)je^{jx}+A(x)j^2e^{jx}+A'x)e^{jx}+A(x)je^{jx}+A(x)e^{jx}= \varepsilon(x)$$ and $A''(x)+(2j+1)A'(x)+(j^2+j+1)A(x)=e^{-jx}\varepsilon(x)$, so $A''(x)+(2j+1)A'(x)=e^{-jx}\varepsilon(x)$. We put $B(x):=A'(x)$. We have $B'(x)e^{(2j+1)x}+(2j+1)e^{(2j+1)x}B(x)=e^{-jx}\varepsilon(x)$, so $\displaystyle B(x)e^{(2j+1)x}=\int_0^xe^{-jt}\varepsilon(t)dt$ and $A'(x)=e^{-(2j+1)x}\int_0^xe^{-jt}\varepsilon(t)dt$, hence we can choose $$A(x)=\int_0^xe^{-(2j+1)s}\int_0^se^{-jt}\varepsilon(t)dtds$$ and finally $$f(x)=C_1e^{jx}+C_2e^{\bar jx}+e^{jx}\int_0^xe^{-(2j+1)s}\int_0^se^{-jt}\varepsilon(t)dtds.$$ Since $\displaystyle\operatorname{Re}j=\frac 12$, we have $\displaystyle\lim_{x\to +\infty}C_1e^{jx}+C_2e^{\bar jx}=0$ and we have to show that the limit of $\displaystyle u(x):=e^{jx}\int_0^xe^{-(2j+1)s}\int_0^se^{-jt}\varepsilon(t)dtds$ is $0$. We have \begin{align*} |u(x)|&\leq e^{-x/2}\int_0^x\int_0^xe^{t/2}|\varepsilon(x)|dtds\\ &=e^{-x/2}\left(\left[s\int_0^se^{t/2}|\varepsilon(x)|dt\right]_{s=0}^{s=x} -\int_0^x se^{s/2}|\varepsilon(s)|ds\right)\\ &=e^{-x/2}\int_0^x(x-t)e^{t/2}|\varepsilon(t)|dt\\ &=\int_0^xse^{-s/2}|\varepsilon(x-s)|ds. \end{align*} Now, we fix $\delta>0$. Let $x_0$ such that for all $x\geq x_0$ we have $|\varepsilon(x)|\leq \delta$. For $x\geq x_0$ we have \begin{align*} |u(x)|&\leq \int_0^{x-x_0}se^{-s/2}|\varepsilon(x-s)|ds+ \int_{x-x_0}^xse^{-s/2}|\varepsilon(x-s)|ds\\ &\leq \delta \int_0^{x-x_0}se^{-s/2}ds+e^{-x/2}\int_0^{x_0}e^{t/2} |\varepsilon(t)|dt\\ &\leq \delta \int_0^{+\infty}se^{-s/2}ds+e^{-x/2}\int_0^{x_0}e^{t/2} |\varepsilon(t)|dt. \end{align*} We get that $\displaystyle\limsup_{x\to +\infty}|u(x)|\leq \delta \int_0^{+\infty}se^{-s/2}ds$ for all $\delta>0$ hence $\displaystyle\lim_{x\to +\infty}u(x)=0$ and we are done.
I guess there is a more elegant way. We note that we can show that if $f\in\mathcal C^1(\mathbb R)$ and $\displaystyle\lim_{x\to +\infty}f(x)+f'(x)=0$ then $\displaystyle\lim_{x\to +\infty}=0$. We use a similar method and the computation is simpler. Note that we can have a result of the form "If $f\in\mathcal C^n(\mathbb R)$ and $\displaystyle\lim_{x\to +\infty}\sum_{k=0}^nf^{(j)}(x)=0$ then $\displaystyle\lim_{x\to +\infty}f(x)=0$" for $n\geq 3$. Indeed, we have that $\displaystyle f(x):=\exp(e^{\frac{2\pi i}n}x)$ is a solution of $\sum_{k=0}^nf^{(j)}(x)=0$, but since $\operatorname{Re}e^{\frac{2\pi i}n}\geq 0$ we can't have $\displaystyle\lim_{x\to +\infty}f(x)=0$.