Prove: $\lim_{x \to \infty} f'(x) = 2$ indicate $\lim_{x \to \infty}f(x) = \infty$ This question is from last year exam in calculus.
In the answers they mentioned more than one way to solve this question.
The method they used is to build a new function but it was a bit complicated.
I would like to get some help.
 A: By definition of limit, let $M$ such that $x \geq M \implies f'(x) > 2-\frac{1}{2} = \frac{3}{2}$. Then we have, for $x > M$
$$f(x) - f(M) = \int_x^M f'(t) dt \geq \frac{3}{2} (x-M),$$
that is
$$ f(x) \geq f(M) + \frac{3}{2} (x-M).$$
Now for any number $L$ we can find $N$ big enough so that
$f(M) + \frac{3}{2} (x-M) > L$
when $x > N$, and this proves that $f(x) \to +\infty$ as $x \to +\infty$.
A: The key is to show that $f$ is bounded below by a function which goes to infinity as $x \to \infty$.  It follows from your hypothesis that there exists an $x_0$ such that for all $x \geq x_0$, $f'(x) \geq 1$.
Now let $x$ be any number bigger than $x_0$.  Consider the average rate of change
$$\frac{f(x) - f(x_0)}{x-x_0}$$
between $x$ and $x_0$.  By the mean value theorem, there exists a $c > x_0$ such that
$$\frac{f(x) - f(x_0)}{x-x_0} = f'(c)$$
Since $f'(c) \geq 1$, this implies that there is a constant $K$ such that $f(x) \geq x +K$ for all $x > x_0$.
A: By L'Hôpital's rule
$$
\mathop {\lim }\limits_{x \to  + \infty } \frac{{f(x)}}{x} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{f'(x)}}{1} = 2.
$$
Thus
$$
\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } x\mathop {\lim }\limits_{x \to  + \infty } \frac{{f(x)}}{x} =  + \infty .
$$
