Show that $f'' > 0$, $\lim_{x \to b^-} = \infty$ implies that $\lim_{x \to b^-} f'(x) = \infty$ Let $f$ be a continuous function on $[a,b)$, $f$ twice differentiable in $(a,b)$ so that $f''(x)>0$ for each $x \in  (a,b)$. Prove that if $$\lim_{x\to b-}f(x) =\infty $$ then 
$$ \lim _{x\to b-}f'(x)=\infty $$
 A: Fix $c$ with $a<c<b$
Then for $c\le x<b$,
\begin{equation}f(x)=f(c)+\int_c^xf'(x)dx\end{equation}
by fundamental theorem of calculus. If $f'(x)$ tended to a finite limit, then $f'(x)$ would have been bounded in a neighbourhood $(b-\delta,b)$. Also, $f'$ being derivable is continuous and will be bounded on the compact interval $[a,b-\delta]$. Hence $f'$ is bounded on $[a,b)$. Let $M$ be a bound for $f'$. Hence for $x\ge c$, $f(x)\le f(c)+(b-c)M$.
Again, since $f$ is continuous on the compact interval $[a,c]$ it is bounded there. This shows that $f$ is bounded on $[a,b)$ contrary to the hypothesis that $f$ blows up as $x$ gets arbitrarily close to $b$.
Any fallacies?

Let me add concluding statement.
Hence the assumption of $\lim_{x\rightarrow b^-}f'(x)<\infty$ was wrong. Hence the limit of $f'$ is infinity.
A: As $f(x) \to \infty$ as $x \to b^-$, we know there exist some $b_i$ in $(a,b)$ s.t. for all $x \in (b_i, b)$, $x \geq i$. We also know that $f$ is continuous, and that the derivative is always increasing. So some ways to proceed would be either of the following:


*

*By considering, say, $\{10^n\}_{n \in \mathbb{N}}$, and the corresponding $b_{10^n}$, you can just use the mean value theorem.

*Suppose not. $f'(x)$ is monotone increasing, and thus if $a_n = b - 1/2^n$ or something similar, the sequence $f'(a_n)$ will have a limit. By our assumption, it won't be infinite. But then you can bound $\lim_{x \to b^-} f(x)$.
Or you could use the MVT in some other way. I suspect that there are many different approaches to this problem relying on the MVT (and the fact that $f'(x)$ is increasing).
A: Suppose $|f'|$ is bounded around $b$. Then for any $x$ and $y$ close to $b$,
$$
|f(x)-f(y)| \leq \left( \sup |f'| \right) |x-y|,
$$
and therefore $\lim_{x \to b-} f(x)$ exists and is finite. This contradiction implies that $|f'|$ is unbounded around $b$. But $f'$ is monotonically increasing, and therefore $f'(x) \to  +\infty$ as $x \to b^{-}$.
A: Because of the notation $b^-$ I presume that $b$ must be finite. Suppose 
$$
m:=\lim_{x \to b^-}f'(x)<\infty.
$$
Since $f''>0$ we deduce that $f'$ is increasing on $(a,b)$. Therefore
$$
f'(x)< m \quad \forall \ x \in (a,b).
$$
Thus for every $x \in [a,b)$ we have
\begin{eqnarray}
f(x)&=&f(a)+\int_a^xf'(t)dt=f(a)+(x-a)m+\int_a^x(f'(t)-m)dt\cr
    &\le& f(a)+(x-a)|m|\le f(a)+|m|(b-a).
\end{eqnarray}
It follows that
$$
\lim_{x \to b^-}f(x) \le f(a)+|m|(b-a),
$$ 
which contradicts the fact that $f(x) \to \infty $ as $x \to b$.
