$\lim\limits_{x\to \infty} f'(x)=1$ implies $f:\mathbb {R\to R}$ is unbounded. Is my solution to the following problem correct?This is a problem from TIFR GS $2014$.Can someone provide me with a better solution?
Let $f:\mathbb {R\to R}$ be differentiable and $f'(x)\to 1$ as $x\to \infty$,show that $f$ is unbounded.
Solution:   Since, $\lim\limits_{x\to \infty} f'(x)=1$,so there exists $x_0\in \mathbb R$ such that,
$x>x_0\implies 1-\frac{1}{100}<f'(x)<1+\frac{1}{100}$
So,$ f'(x)>0$ for all $x\in [x_0,\infty)$
So,$f$ is increasing on $[x_0,\infty)$.
If $f$ is bounded,then let $s=\sup\limits_{x\in [x_0,\infty)} f(x)$.
Let $\epsilon>0$ (small)
Take $s-\epsilon$,then there exists $x\in [x_0,\infty)$ such that $s-\epsilon<f(x)\leq s$.
Since,$f$ is increasing on $[x_0,\infty)$,so for each $t>x$ we have $s-\epsilon<f(x)\leq f(t)\leq s$.
Choose $x+2\epsilon>x$
We can apply LMVT on $f$ in the interval $[x,x+2\epsilon]$ to get $c\in (x,x+2\epsilon)$ such that $f(x+2\epsilon)=f(x)+2\epsilon . f'(c)$
Now,$1-\frac{1}{100}<f'(c)<1+\frac{1}{100}$
So,$2\epsilon(1-\frac{1}{100})<2\epsilon.f'(c)<2\epsilon(1+\frac{1}{100})$
$\implies f(x)+2\epsilon(1-\frac{1}{100})<f(x+2\epsilon)<f(x)+2\epsilon(1+\frac{1}{100})$
Now,$s-\epsilon<f(x)\leq s$,so $s<s+\frac{49\epsilon}{50}=s+\epsilon-\frac{\epsilon}{50}=s-\epsilon+2\epsilon(1-\frac{1}{100})<f(x)+2\epsilon(1-\frac{1}{100})<f(x+2\epsilon)$,which is a contradiction as $f(x+2\epsilon)\leq s$.
 A: The idea is that $f'$ is eventually almost 1 and hence it grows like $g(x)=x$. 
To make it more precise, since $f'(x)\to 1$ as $x\to \infty$ there is $M>0$ such that,
\begin{equation}
f'(x)> \frac12,
\end{equation}
for all $x>M$. By the mean value theorem, for each $x>M$ there is $M< \xi< x$ such that;
\begin{equation}
\frac{f(x)-f(M)}{x-M}= f'(\xi)> \frac12.
\end{equation}
This implies that,
\begin{equation}
f(x)\geq f(M)+ \frac12(x-M),
\end{equation}
for all $x\geq M$. Let $x\to \infty$ and you get the result.
A: This is one place where usage of L'Hospital's Rule actually makes sense (but students would forget it because they think it can be used only for limit evaluation).
Since $f'(x) \to 1$ as $x\to\infty $ it follows by L'Hospital's Rule that $f(x) /x\to 1$ as $x\to\infty $. Thus $f(x) $ is unbounded as $x\to\infty $.
Without L'Hospital's Rule you can try this approach also. Since $f'\to 1$ the derivative is positive in some interval of type $[a, \infty) $ and hence $f$ is strictly increasing in this interval. Thus either $f(x) $ tends to a finite limit $L$ or to $\infty $ as $x\to\infty $. But if $f(x) \to L$ then by mean value theorem $$f(x+1)-f(x)=f'(c)$$ and we get a contradiction as left hand side tends to $L-L=0$ and right hand side tends to $1$. It follows that $f(x) \to\infty $ as $x\to\infty $. 
A: You mean,just to show unboundedness?Yes,it can be done pretty easily.Observe that there exists $M>0$, such that for all $x\geq M$ we have $f^{\prime }(x)\geq \frac{1}{2}$.Thus,for all $x>M$,using mean value theorem,we have $f(x)\geq \frac{(x-M)}{2}+ f(M)$, so $f$ is bounded below by a linear map which itself is unbounded.We can of course, sandwich $f^{\prime}$ eventually between $\frac{1}{2}$ and $\frac{3}{2}$ and using a similar approach, deduce that $f(x)=\Theta(x)$, that is, the growth of $f$ is linear.
Deleted part : I initially misread your question and took $f$ to be twice differentiable.I mean,what can you infer about the asymptotic behaviour of $f$ when $f^{\prime \prime}$ tends to $1$ as $x\to \infty$ ?In this case,as you see,we have a pretty quick solution too.We can eventually bound $f^{\prime \prime}$ from below by $\frac{1}{2}$ and then use the fact that eventually convex maps on $\mathbb{R}$ are unbounded,a proof of which can be pretty geometrically pleasant.In this case too,like the previous one,you can prove $f^{\prime \prime}(x)=\Theta(x^2)$.In general,the more regularity you assume on $f$ the higher the degree of the asymptotic polynomial bound gets.
P.S   I was asked the above question with second derivatives at the NBHM Ph.D scholarships interview this year.
