If $f'$ tends to a positive limit as $x$ approaches infinity, then $f$ approaches infinity Some time ago, I asked this here. A restricted form of the second question could be this:

If $f$ is a function with continuous first derivative in $\mathbb{R}$ and such that $$\lim_{x\to \infty} f'(x) =a,$$ with $a\gt 0$, then $$\lim_{x\to\infty}f(x)=\infty.$$

To prove it, I tried this:
There exist $x_0\in \mathbb{R}$ such that for $x\geq x_0$,
$$f'(x)\gt \frac{a}{2}.$$
There exist $\delta_0\gt 0$ such that for $x_0\lt x\leq x_0+ \delta_0$
$$\begin{align*}\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)&\gt -\frac{a}{4}\\
\frac{f(x)-f(x_0)}{x-x_0}&\gt f'(x_0)-\frac{a}{4}\\
&\gt \frac{a}{2}-\frac{a}{4}=\frac{a}{4}\\
f(x)-f(x_0)&\gt \frac{a}{4}(x-x_0)\end{align*}.$$
We can assume that $\delta_0\geq 1$. If $\delta_0 \lt 1$, then $x_0+2-\delta_0\gt x_0$ and then $$f'(x_0+2-\delta_0)\gt \frac{a}{2}.$$
Now, there exist $\delta\gt 0$ such that for $x_0+2-\delta_0\lt x\leq x_0+2-\delta_0+\delta$ $$f(x)-f(x_0+2-\delta_0)\gt \frac{a}{4}(x-(x_0+2-\delta_0))= \frac{a}{4}(x-x_0-(2-\delta_0))\gt \frac{a}{4}(x-x_0).$$ It is clear that $x\in (x_0,x_0+2-\delta_0+\delta]$ and $2-\delta_0+\delta\geq 1$.
Therefore, we can take $x_1=x_0+1$. Then $f'(x_1)\gt a/2$ and then there exist $\delta_1\geq 1$ such that for $x_1\lt x\leq x_1+\delta_1$ $$f(x)-f(x_1)\gt \frac{a}{4}(x-x_1).$$
Take $x_2=x_1+1$ and so on. If $f$ is bounded, $(f(x_n))_{n\in \mathbb{N}}$ is a increasing bounded sequence and therefore it has a convergent subsequence. Thus, this implies that the sequence $(x_n)$:
$$x_{n+1}=x_n+1,$$ have a Cauchy's subsequence and that is a contradiction. Therefore $\lim_{x\to \infty} f(x)=\infty$.
I want to know if this is correct, and if there is a simpler way to prove this. Thanks.
 A: There is a simpler way to prove this. (I apologize but I have not looked at your solution; perhaps another user will comment in this regard.) Let $M$ be a positive integer. Choose a positive integer $N$ such that $x>N$ implies $f'(x)>\frac{a}{2}$. If $y>\text{max}\{N,\frac{2(M-f(N))}{a}+N\}$, then $y>N$ and the mean value theorem implies that $f(y)-f(N)=f'(x_y)(y-N)$ for some real number $N<x_y<y$. Furthermore, $f(y)=f(N)+f'(x_y)(y-N)>f(N)+\frac{a(y-N)}{2}>M$ for all such $y$. Since $M$ was arbitrary, it follows that $\lim_{x\to\infty} f(x)=\infty$.
A: After reading some of the other answers, I thought it might be helpful to rephrase Shai Covo's answer in more physical language: suppose that instead of $y = f(x)$ we have $x = x(t)$, the position function of a particle at time $t$.  (Of course it makes no mathematical difference what we call the variables...)
The hypothesis is that the velocity function $x'(t)$ approaches a limiting value $v_0 > 0$ as $t$ approaches infinity.  Therefore there must exist a $t_0 \in \mathbb{R}$ such that for all $t \geq t_0$, $v(t) \geq \frac{v_0}{2} > 0$.  Let us therefore compare the position function $x(t)$ to the following one:
$$y(t) = \frac{v_0}{2}\left(t-t_0\right) + x(t_0).$$
We observe:
1) $y(t)$ is a linear function with positive slope, so certainly $\lim_{t \rightarrow \infty} y(t) = \infty$.  
2) We have $x(t_0) = y(t_0)$ and $x'(t) \geq y'(t)$ for all $t \geq t_0$.
Thus on the time interval $[t_0,\infty)$, the particle $x = x(t)$ starts out at the same position as the particle $y = y(t)$ and for all times $t$ the instantaneous velocity of the first particle is always greater than or equal to that of the second particle.  Therefore physical intuition tells us that we must have $x(t) \geq y(t)$ for all $t \geq t_0$, and hence $\lim_{t \rightarrow \infty} x(t) = \infty$.  
How do we make this rigorous?  Replacing $x(t)$ and $y(t)$ by $x(t) - y(t)$ and $0$, what we need to show is that a function which is non-negative at a point $t_0$ and has everywhere non-negative derivative is itself non-negative for all $t \geq t_0$.  But now we can take our pick of mathematical justifications for this: if we assume that the derivative is continuous (as the OP does) then $x(t) = \int_{t_0}^t x'(t) dt$ and the integral of a non-negative function is certainly non-negative, being a limit of non-negative Riemann sums.  Or indeed, if we assume that there exists $t_1 > t_0$ such that $x(t_1) < x(t_0)$ and apply the Mean Value Theorem, then we get 
$$x(t_1) - x(t_0) = x'(c) (t_1-t_0).$$
The left hand side is negative and the right hand side is non-negative: contradiction!
In summary, reasoning physically in a mathematically careful way reduces us to a statement which is easy to prove mathematically.
As a side remark, some educators have suggested that this special case of the Mean Value Theorem  (called the Mean Value Inequality, Increasing Function Theorem, and so forth) is indeed more intuitive than the full Mean Value Theorem and have proposed developing calculus around it.  In my experiences teaching calculus I have found it helpful to point out explicitly this special case of MVT, but that's about it: the full MVT is also (more) useful and not much more bitter a pill...
A: Since you assume that $f$ has first derivative continuous on $\mathbb{R}$, there is an alternative to the Mean-value theorem, namely the Fundamental theorem of calculus (the former is very useful for cases when $f'$ is not assumed continuous). So, from $\lim _{x \to \infty } f'(x) = a > 0$, we have that there exists $M > 0$ such that $f'(x) > a/2$ for any $x \geq M$. Since $f'$ is assumed continuous, we can apply the the Fundamental theorem of calculus to obtain
$$
f(x) - f(M) = \int_M^x {f'(u)\,du}  \ge \int_M^x {\frac{a}{2}\,du}  = \frac{a}{2}(x - M),
$$
for any $x > M$.
Since the right-hand side goes to $\infty$ as $x \to \infty$ (note that $M$ is fixed), so does $f(x) - f(M)$. Hence $
\lim _{x \to \infty } f(x) = \infty $.
EDIT: The above was just to give an alternative to the the Mean-value theorem (used in Amitesh Datta's answer); its  analog for the Mean-value theorem is as follows.
From $\lim _{x \to \infty } f'(x) = a > 0$, we have that there exists $M > 0$ such that $f'(x) > a/2$ for any $x \geq M$. For any $x > M$, since $f$ is continuous on the closed interval $[M,x]$ and differentiable on the open interval $(M,x)$, by the Mean-value theorem there exists $c = c(M,x) \in (M,x)$ such that
$$
f(x)-f(M)=f'(c)(x-M).
$$
Since $c \geq M$, $f'(c) > a/2 $, and hence
$$
f(x)-f(M) \geq \frac{a}{2}(x - M).
$$
The conclusion is as before. 
Note: As already indicated above, the proof using the Mean-value theorem has the advantage that it works for any differentiable function $f: \mathbb{R} \to \mathbb{R}$.
A: HINT  $\  \  $ Writing $\rm\ f\ =\ (f/x)\ x\ \ $ yields a determinate limit (using L'Hôpital's rule), namely
$$\rm\displaystyle\quad \lim_{x\to\infty}\ f\ =\ \lim_{x\to\infty} \dfrac{f}x\ \lim_{x\to\infty}\ x\ =\ \lim_{x\to\infty}\dfrac{f\:\:'}{1\:\ }\ \lim_{x\to\infty}\ x\ =\ a\ \cdot \infty\ =\ \infty\quad by\quad a > 0$$
NOTE $\ $ Below is said L'Hospital's rule, from Rudin's $\:$ Principles of Mathematical Analysis, $\:$  1976. Above I use the general form $(15)$ of the  $\rm \infty/\infty$ rule. It requires only that the denominator $\to\infty\:.$ For further such generalizations see the Monthly papers cited here.

A: I will try to prove is in a different way which can be much simpler - using visualization.  
Imagine how will a function look if it has a constant, positive slope -
A straight line, with a positive angle with the positive x axis.
Although this can be imagined, I am attaching a simple pic -

(Plot of our imaginative function - $f(x)$ vs $x$)  
As per the situation in the question, for $f(x)$ the slope exists (and is finite) at all points, so it means that the function is continuous. Since the slope is also constant at $\infty$, $f$ has to be linear at $\infty$. Thus, the graph of the function should be similar to the above graph.

(assume the value of x to be as large as you can imagine.)  
Hence, putting the above situation mathematically, we have,  
If $\lim_{x\to \infty}\ f'(x) =a\qquad(with\ a>0)$
then $\lim_{x\to\infty}\ f(x)=\infty.$
