Prove that $\lim_{x \to \infty} f(x) = \infty $ 
*

*Let $f: \mathbb R \to \mathbb R$ be a function such that there exists constants $ b,\ m> 0,\ c \in \mathbb R $ such that 
$$f(x) > mx+c,\  \forall x>b$$
Prove that $\lim_{x \to \infty} f(x) = \infty $

*If $f$ is a differentiable function on $(0, \infty)$ and $f,\ f'$ both are strictly increasing on $(0, \infty )$, prove that $\lim_{x \to \infty} f(x) = \infty $. Use mean value theorem and the previous result.
I did the first part as follows- 
Let $a>0$
Define $p:=\max \{b, \dfrac{|a-c|}{m}\}$
\begin{align*}
x>p &\implies x>b\ \text{and}\ x>\frac{|a-c|}{m} \\
x>b &\implies f(x)> mx+c \\
x> \dfrac{|a-c|}{m} &\implies mx+c > a > c-mx \\
\therefore x>p &\implies f(x) > mx+ c > a (>0) \\ 
\forall a>0, \exists p \in \mathbb Z_+ st, x>p\ &\implies |f(x)|>a\\
\therefore \lim_{x \to \infty} f(x) &= \infty\\
\end{align*}
I'm stuck on the second part, any help?
 A: Since $f$ is strictly increasing, $f'$ is positive on $(0,\infty)$. Let $m=f'(1)>0$.
Also, since $f'$ is strictly increasing, $f'>m$ on $(1,\infty)$.
If $x$ is an arbitrary point in $(1,\infty)$, by the Mean Value Theorem(we can use this due to the condition that $f$ is differentiable, which says that $f$ is continuous.), there exists a $c\in(1,
x)$ such that
$${f(x)-f(1)\over x-1}=f'(c)>m.$$
Thus $f(x)>mx+(f(1)-m)=mx+c$, where $c$ is a constant.
Therefore, by the result of 1, $\displaystyle \lim_{x\to\infty}f(x)=\infty$.
A: A possible solution could be as follows: Fix $a,b$ with $0<a<b$. According to the mean value theorem we have that exists a $\lambda$ between $a$ and $b$ ($a<\lambda<b$) such that $\displaystyle\frac{f(b)-f(a)}{b-a}=f'(c)>0$ because we already know that $f(b)>f(a)$.
Now we construct the tangent line to the point $(c,f(c))$ in order to use the part one of the problem. As you already know, the tangent line is given by $y-f(c)=f'(c)(x-c)$ which means that $y=xf'(c)+f(c)-cf'(c)=mx+\beta$ if we call $m=f'(c)$ and $\beta=f(c)-cf'(c)$.
Part one says that it should be enought to prove that $f(x)>mx+\beta$ for $x>b$. Let's see if we can do that: take $k>b$,  the inequality we have to prove is $f(k)>kf'(c)+f(c)-cf'(c)$, or which is the same, prove $f(k)-f(c)>f'(c)(k-c)$ and the last inequality can be written as $\displaystyle\frac{f(k)-f(c)}{k-c}>f'(c)$ where the left-hand side looks like an obvious form of the mean value theorem.
Now, applying the mean value theorem again to the left hand side, exists $u$ with $c<u<k$ such that $\displaystyle\frac{f(k)-f(c)}{k-c}=f'(u)$. Before we used that $f$ was an increasing function, now we use that  $f'$ is also increasing, therefore, $f'(c)<f'(u)$, which proves $\displaystyle\frac{f(k)-f(c)}{k-c}>f'(c)$ and thus and $f$ is always above the tangent line to $(c,f'(c))$ when $x>b$.
The limit follows from what you already proved.
