If $f$ is continuous on $[a,b)$ and differentiable on $(a,b)$ such that $\lim_{x\to b^{-}}f(x)=\infty$, Then $f'$ is not bounded above in $(a,b)$. I got this problem:
Let $f$ be a continuous function on the interval $[a,b)$ and differentiable on the interval $(a,b)$, Prove that if $\lim_{x\to b^{-}}f(x)=\infty$, Then $f'$ is not bounded above in $(a,b)$.
I tried some ways but none led me to a solution.
Any help will be appreciated.
 A: Let show that if $f'$ is bounded on $(a,b)$ then $f$ is bounded on $(a,b)$ too.
Step 1: $f$ is uniformly continuous on $(a,b)$.
We suppose of course that $f\in\mathcal C^1(a,b)$. By hypothesis, since $f'$ is bounded, 
$$|f'|\leq M$$ for a certain $M>0$. By the mean value theorem,
$$|f(x)-f(y)|\leq M|x-y|$$
for all $x,y\in(a,b)$. Then $f$ is uniformly continuous on $(a,b)$. Indeed, if $\varepsilon>0$ is fix and $\delta=\frac{\varepsilon}{M}$, if $|x-y|<\delta$ then, 
$$|f(x)-f(y)|<\varepsilon$$
for all $x,y\in(a,b)$.
Step 2: $f$ is bounded on $(a,b)$.
Let show that $\lim_{x\to b^-}f(x)$ exist. Let take a sequence $(a_n)_{n\in\mathbb N}\subset (a,b)$ such that $\lim_{n\to\infty }a_n=b$ and let show that $\left(f(a_n)\right)_{n\in\mathbb N}$ is a Cauchy sequence. Let $\varepsilon>0$. By uniform continuity of $f$, there is a certain $\delta>0$ such that $$|x-y|<\delta\implies |f(x)-f(y)|<\varepsilon.$$Clearly, $(a_n)_{n\in\mathbb  N}$ is a Cauchy sequence. Then, if $n,m>N$, $$|a_n-a_m|<\delta$$
for a certain $N\in\mathbb  N$, and so, if $n,m>N$,
$$ |f(a_n)-f(a_m)|<\varepsilon.$$
Then we have proved that $\left(f(a_n)\right)_{n\in\mathbb N}$ is a Cauchy sequence, and so it converge. The last thing that we have to prove is that for all $(b_n)\subset (a,b)$ such that $b_n\to b$ if $n\to\infty $, the limits of the sequences $\left( f(b_n)\right)_{n\in\mathbb  N}$ are the same. Suppose that it's not. Then there exists two sequences $(b_n)_{\in \mathbb  N},(c_n)_{n\in\mathbb  N}\subset (a,b)$ such that $$\lim_{n\to \infty }b_n=\lim_{n\to\infty }c_n=b$$ and $$\lim_{n\to\infty }f(b_n)=:\ell_1\neq \ell_2:=\lim_{n\to\infty }f(c_n).$$
Now let define $d_n=b_n$ if $n$ is even and $d_n=c_n$ if $n$ is odd. Clearly, $$\lim_{n\to\infty }d_n=b$$ but $$\lim_{n\to\infty }f(d_n)$$ doesn't exist which is a contradiction with the fact that for all sequence $(b_n)_{n\in\mathbb N}$ such that $\lim_{n\to\infty }b_n=b$, the sequence $\left (f(b_n)\right)_{n\in\mathbb  N}$ converge. As well, we prove that $\lim_{x\to a^-}f(x)$ exist. And that conclude that $f$ is bounded on $[a,b]$ and so on $(a,b)$.
This finally prove that $f'$ is bounded on $(a,b)$ imply that $f$ is bounded on $(a,b)$, and so if $f$ is unbounded, $f'$ is unbounded too.
Q.E.D.
A: Hint: it's tempting to use the fundamental theorem of calculus – it was my first thought. But you don't know that $f'$ is continuous, so it might not be integrable. So try the mean value theorem instead?
A: Assume that the derivative $f'$ remains bounded in a (left) neighborhood of $b$: $|f'| \leq M$. Now, pick $\varepsilon>0$ and choose any $x$, $y$ so close to $b$ that $|f'| \leq M$ in the interval $[x,y]$. Therefore, by the mean value theorem, for some $\xi$ between $x$ and $y$,
$$
|f(x)-f(y)| = |f'(\xi)| |x-y| \leq M |x-y| < \varepsilon
$$
provided that $|x-y|<\varepsilon/M$. This tells us that $f(x)$ and $f(y)$ can be make as close as we wish, provided that we choose $x$ and $y$ sufficiently close to $b$. Hence $\lim_{x \to b-} (x)$ exists in $\mathbb{R}$, contradicting the assumption of the exercise.
