If $f$ is convex (or concave), does the existence of $\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x}$ imply the existence of asymptote?

Question

A follow-up question for the previous one.

To compute the oblique asymptote as $x \to +\infty$, we can first compute $\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x}$, if it exists, and $\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = k$, then we can further compute $\mathop {\lim }\limits_{x \to + \infty } (f(x) - kx)=b$, and if it exists then the asymptote would be $y = kx + b$.

Now the existence of $\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = k$ does not in general imply existence of the second limit $\mathop {\lim }\limits_{x \to + \infty } (f(x) - kx)=b$, as a counterexample $f(x)=x+\sin x$ shown by levap in the previous question.

But what if we restrict $f$ being convex (or concave)? Will the claim hold in this case? The counterexample $f(x)=x+ \sin x$ now fails as it is neither convex or concave.

Answer 1

As others (user Michael, who for some reason did not believe in himself) noticed, there is the concave counterexample with $k=0$ $$f(x)=\begin{cases} \ln x&\text{if }x>1\\ x-1&\text{if }x\le1\end{cases}$$

And hence the convex counterexample $-f(x)$ with $k=0$. And $-f(x)+\alpha x$ for arbitrary $k$.