Edit: I forgot to credit user23571113 for their valuable comment on the OP that made this answer possible!
Actually, it turns out that if any two derivatives have limits as $x\to\infty$, that forces all intermediate derivatives to have limits as well. More precisely:
Claim: Let $k\ge2$, and let $f(x)$ be a $k$-times differentiable function. Suppose that $\lim_{x\to\infty} f(x)$ and $\lim_{x\to\infty} f^{(k)}(x)$ both exist. Then $\lim_{x\to\infty} f'(x) = \lim_{x\to\infty} f''(x) = \cdots = \lim_{x\to\infty} f^{(k)}(x) = 0$.
Proof: First we prove that $\lim_{x\to\infty} f^{(k)}(x) = 0$. Suppose that $\lim_{x\to\infty} f^{(k)}(x)$ were positive; then there exists $c>0$ and $x_0$ such that $f^{(k)}(x) > c$ for $x>x_0$. Integrating $k$ times, we conclude that $f(x) > \frac c{k!} (x-x_0)^k + P(x)$ where $P$ is a polynomial of degree at most $k-1$. But this contradicts the assumption that $\lim_{x\to\infty} f(x)$ exists. Therefore $\lim_{x\to\infty} f^{(k)}(x)$ cannot be positive, and a similar argument shows that $\lim_{x\to\infty} f^{(k)}(x)$ cannot be negative either.
Now we prove that $\lim_{x\to\infty} f'(x) = \lim_{x\to\infty} f''(x) = \cdots = \lim_{x\to\infty} f^{(k-1)}(x) = 0$. Let $\varepsilon>0$ be given. Choose $x_0$ such that both $|f(x)-f(x_0)|$ and $|f^{(k)}(x)|$ are less than $\varepsilon$ for all $x>x_0$. For any $u>0$, we can apply Taylor's theorem to $f(x)$ at the point $x+u$:
\begin{align*}
f(x+u) = f(x) + f'(x)u + \frac12f''(x)u^2 + \cdots + \frac1{(k-1)!} f^{(k-1)}(x)u^{k-1} + \frac1{k!} f^{(k)}(y)u^k
\end{align*}
for some $y$ in $(x,x+u)$. Applying this with any $x>x_0$ and with $u=1,2,\dots,k-1$, we see that there exist $y_1,\dots,y_{k-1}>x>x_0$ such that
\begin{align*}
f'(x)1 + \frac12f''(x)1^2 + \cdots + \frac1{(k-1)!} f^{(k-1)}(x)1^{k-1} &= f(x+1) - f(x) - \frac1{k!} f^{(k)}(y_1)1^k \\
f'(x)2 + \frac12f''(x)2^2 + \cdots + \frac1{(k-1)!} f^{(k-1)}(x)2^{k-1} &= f(x+2) - f(x) - \frac1{k!} f^{(k)}(y_2)2^k \\
&\vdots \\
f'(x)(k-1) + \cdots + \frac1{(k-1)!} f^{(k-1)}(x)(k-1)^{k-1} &= f(x+k-1) - f(x) - \frac1{k!} f^{(k)}(y_{k-1})(k-1)^k.
\end{align*}
Note that the right-hand sides are all less than $k^k \varepsilon$ (say) in absolute value. These are $k-1$ linear equations in the quantities $f'(x),\dots,f^{(k-1)}(x)$; moreover, the matrix of coefficients is a Vandermonde matrix and is thus invertible; moreover the entries of its inverse are all bounded in absolute value by some function of $k$. By applying the inverse matrix to both sides of the associated matrix equation, we conclude that when $x>x_0$, each of $f'(x),\dots,f^{(k-1)}(x)$ is itself bounded by $\varepsilon$ times some constant depending only on $k$. Since this is true for any $\varepsilon>0$, we conclude that $\lim_{x\to\infty} f'(x) = \lim_{x\to\infty} f''(x) = \cdots = \lim_{x\to\infty} f^{(k-1)}(x) = 0$ as claimed.
