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 $x_0\lt x\leq x_0+ \delta_0$
$$\begin{align*}\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)&amp;\gt -\frac{a}{4}\\
\frac{f(x)-f(x_0)}{x-x_0}&amp;\gt f'(x_0)-\frac{a}{4}\\
&amp;\gt \frac{a}{2}-\frac{a}{4}=\frac{a}{4}\\
f(x)-f(x_0)&amp;\gt \frac{a}{4}(x-x_0)\end{align*}.$$
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.
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 $f&#39;$ is not assumed continuous). So, from $\lim _{x \to \infty } f&#39;(x) = a &gt; 0$, we have that there exists $M &gt; 0$ such that $f&#39;(x) &gt; a/2$ for any $x \geq M$. Since $f&#39;$ is assumed continuous, we can apply the the Fundamental theorem of calculus to obtain
$$
f(x) - f(M) = \int_M^x {f&#39;(u)\,du}  \ge \int_M^x {\frac{a}{2}\,du}  = \frac{a}{2}(x - M),
$$
for any $x &gt; M$.
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 $\lim _{x \to \infty } f&#39;(x) = a &gt; 0$, we have that there exists $M &gt; 0$ such that $f&#39;(x) &gt; a/2$ for any $x \geq M$. For any $x &gt; M$, 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
$$
f(x)-f(M)=f&#39;(c)(x-M).
$$
Since $c \geq M$, $f&#39;(c) &gt; 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 $f: \mathbb{R} \to \mathbb{R}$.
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 $\lim_{x\to \infty}\ f'(x) =a\qquad(with\ a&gt;0)$

then $\lim_{x\to\infty}\ f(x)=\infty.$