strictly convex functions and limits Suppose I have a strictly convex function $f(x)$ for $x\geq 0$, with $f(0) = 0$, $f'(0) =0$ and $f''(0) >0$.  Is it obvious that $f$ must be superlinear as $x\to +\infty$?  Alternatively, how can I conclude that
\begin{equation}
\lim_{x\to \infty}\frac{f(x)}{x} = +\infty
\end{equation}
 A: You can’t. Consider 
$$f(x)=\sqrt{1+x^2}-1\;;$$
clearly $f(0)=0$. Now
$$f'(x)=\frac{x}{\sqrt{1+x^2}}\;,$$
so $f'(0)=0$, and 
$$f''(x)=\frac{\sqrt{1+x^2}+\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}=\frac{1+2x^2}{(1+x^2)^{3/2}}>0$$
for all $x$, but
$$\lim_{x\to\infty}\frac{f(x)}x=1\;.$$
A: $f(x) = x -\ln (1+x)$ is a counterexample. More generally, let $g:[0,\infty) \to [0,\infty)$ be any strictly increasing continuous function such that $g(0)=0, g'(0) > 0$ and $\lim_{t\to \infty}g(t)<\infty.$ Then $f(x) = \int_0^x g(t)\,dt$ is a counterexample. (That's how I found $x -\ln (1+x);$ it's the integral of $t/(t+1).$)
A: The claim is true for strongly convex function, but not generally true for strictly convex functions. The -ve result for strictly convex functions is proved by @Brian M. Scott's counter-example. For strongly convex functions, indeed one would have
\begin{eqnarray}
\exists m > 0\text{ s.t }f(x) \ge f(0) + f'(0)(x-0) + \frac{m}{2}x^2 = \frac{m}{2}x^2, \forall x
\end{eqnarray}
from which it would follow $\underset{x \rightarrow +\infty}{\text{lim }}\frac{f(x)}{x} = +\infty$.
N.B.: In the above definition of strong convexity, the global constant $m > 0$ is called a modulus of strong convexity for $f$. In @Brian M. Scott's counter-example (as well as any other counter-example) for the strict convexity case, we won't be able to globally lower-bound $f''$ by a positive constant :)
