Show that $\lim_{x \rightarrow \infty} f(x)$ exists by the given condition. Suppose $f$ is a twice differentiable function such that $f''(x) \ge 0$ for all $x \in \mathbb R$. Let $0 \le f(x) \le 1$ for $x \ge 0$. Then show that $\lim_{x \rightarrow \infty} f(x)$ exists.
I have tried it but I fail. Please give me some hint to proceed in the right way.
Thank you in advance. 
 A: $f''>0$ therefore $f'$ is increasing.


*

*First case: $\forall x, f'(x)\le0$. Then $f$ is always decreasing, and since it's bounded it converges.

*Second case: Likewise, $\forall x, f'(x)\ge0$. Then $f$ is always increasing, and since it's bounded it converges.

*Third case: $f'$ is negative until some $x_0$ where it becomes positive (remember, $f'$ is increasing). Then we're back to case 2 and $f$ converges.
Edit:
As pointed out by @MatthewLeingang, cases 2 and 3 are not actually necessary. Indeed, a function that verifies $f''(x)>0,f'(x)>0$ for all $x\ge x_0$ cannot converge.
This can be proved easily: suppose that $\lim_{x\to\infty}f(x)=l<\infty$. By definition for any $\epsilon>0$ there exists $x_1\ge x_0$ such that $|l-f(x_1)|=l-f(x_1)<\epsilon$. ($l>f(x_1)$ since $f$ is increasing)
However, since $f'$ is also increasing, we have $$f(x_1+\frac{2\epsilon}{f'(x_1)})\ge f(x_1)+2\epsilon>l$$ which is absurd. Hence $f$ does not converge.
A: A brief answer.
If $f:\mathbb R\to\mathbb R$ is convex, then exactly one of the following three holds:
a. $f$ is increasing,
b. $f$ is decreasing,
c. There exists an $x_0\in\mathbb R$, such that $f$ is decreasing in $(-\infty,x_0]$ and increasing in $[x_0,\infty)$.
In all three cases, $f$ is eventually monotonic (i.e., it is monotonic in some interval $[a,\infty)$, for some $a\ge 0$). Since $f$ is also bounded, in the same interval, then the limit $\lim_{x\to\infty}f(x)$ exists, and it lies in the interval $[0,1]$, since $0\le f(x)\le 1$, for $x\ge 0$.
