Prove: $\lim_{x \to \infty} f'(x) = 2$ indicate $\lim_{x \to \infty}f(x) = \infty$

Question

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.

Answer 1

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 $x \to +\infty$.

Answer 2

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 $x > x_0$.

Answer 3

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 $$ \mathop {\lim }\limits_{x \to + \infty } f(x) = \mathop {\lim }\limits_{x \to + \infty } x\mathop {\lim }\limits_{x \to + \infty } \frac{{f(x)}}{x} = + \infty . $$