# 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.

• To clarify, are you asking to prove $\lim_{x \to \infty} f'(x) = 2 \implies \lim_{x \to \infty} f(x) = \infty$? Commented Mar 1, 2021 at 14:14
• This should be obvious intuitively: eventually $f$ is increasing at a rate of around $2$ vertical units per horizontal unit. So it's going to get bigger forever.
– D_S
Commented Mar 1, 2021 at 14:15
• Welcome to Mathematics Stack Exchange! By the way, I think that the only value of $a$ for which $\lim_{x\to\infty}f'(x)=a$ does not imply $\lim_{x\to\infty}f(x)=\pm\infty$ is $a=0$. Commented Mar 1, 2021 at 14:18
• Hint: Use Mean Value Theorem on each intervals $[n,n+1]$ for $n \in \Bbb N$ and then do a telescoping sum Commented Mar 1, 2021 at 14:19

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 $$x \to +\infty$$.

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 $$x > x_0$$.

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

• How can you apply l'Hospital without knowing à priori that you are in a $\frac \infty\infty$ form ? This is circular proof.
– zwim
Commented Mar 1, 2021 at 15:19
• @ zwim LH applies to limits of the form ${{\rm anything}\over\infty}$. Commented Mar 1, 2021 at 15:48