The answer is probably different, depending on what you mean by the limit.
Complex limits. The answer seems to be yes. What follows is a heuristic argument. Elementary functions are analytic, with at most countably many singularities. So when the finite limit exists, the singularity is removable or a ramification point. Expanding everything in power series (perhaps with fractional powers), we can compute the limit. The coefficients of these power series are elementary constants, because they are expressed in terms of derivatives. So the answer is probably yes, if we mean complex limits. (This is heuristic because singularities can accumulate so one needs a bit more careful argument).
Real limits. The answer seems to be no. Consider a trigonometric sum $$f(t)=\sum_{j=1}^n a_j\exp(\lambda_jit),\quad i=\sqrt{-1},$$ and assume that it has no real zeros. Suppose that the $\lambda_j$ are real "elementary constants" but incommensurable. Then $f(t)=r(t)\exp(i\phi(t))$, where $\phi$ is a well-defined elementary real function. The limit $$m:=\lim_{t\to+\infty}\phi(t)/t$$ always exists: this is the celebrated Mean Motion Theorem. There is a formula for this limit due to A. Wintner: $$m=\sum\lambda_jW_j,\quad\mbox{where}\quad W_j=\int_{T^n}\Re\frac{a_j\exp(i\theta_j)}{\sum_ka_k\exp(i\theta_k)}d\theta_1\ldots d\theta_n,$$ where $T^n=[0,2\pi]^n$. These integrals are probably not "elementary constants": they can be expressed in Bessel functions. Again, this argument is incomplete, because one has to prove that the $W_j$ can indeed be "non-elementary" constants.