Asymptotic behavior of $|f'(x)|^n e^{-f(x)}$

Let $f$ be a strictly convex function on $\mathbb R$, $f'' \geq C > 0$. Let $n$ be a positive integer. What can we say about the growth rate of $|f'(x)|^n e^{-f(x)}$ as $x\rightarrow \infty$? Must it go to $0$? If so, how rapidly? Intuitively, it seems like it should at least tend to $0$, as $e^{-f(x)}$ will decay at a very rapid rate, and $f'$ cannot grow too much more rapidly than $f$ does. Does anyone have any suggestions as to how to prove this, or any counterexamples?

No, $|f'|^n e^{-f}$ need not go to $0$. We can have $\limsup_{x\to\infty} |f'(x)|^n e^{-f(x)} = \infty$.

Let $$f(x) = \frac{C}{2} x^2 +\sum_{k=0}^\infty (\max(x-k,0))^{p_k}$$ where $p_k \ge 3$ are to be chosen inductively. Note that $f''$ exists everywhere and $f''\ge C$. For an integer $m\ge 1$, the value $f(m)$ depends only on $p_k$ with $k\le m-2$, because the term with $k=m-1$ contributes $1$ to the sum and subsequent terms contribute $0$. On the other hand, $f'(m)$ consists of the terms that depend on $p_k$ with $k\le m-2$, plus $p_{m-1}$. Therefore, we can choose $p_{m-1}$ to make, e.g.,
$$f'(m)> \exp(\exp(\exp(\exp( f(m)))))$$ and continue to do this for every $m$.