Limit and derivative

Question

The function $f:\Bbb R\to\Bbb R$ is differentiable and $\lim_{x\to\infty}f'(x)=1$. Which of the following is true?

A) $f$ is bounded
B) $f$ is increasing
C) $f$ is unbounded
D) $f'(x)$ is bounded

Answer 1

A. This suggests that $\displaystyle \lim_{x \to \infty} f(x) = x+K$, for some constant $K$, so not bounded, because it increases forever.

B. An increasing function is ALWAYS increasing. We have no evidence that states that this function is monotone.

C. If A is false, then C is true.

D. $f'(x)$ must exist for all $x \in \mathbb R$. It is also bounded to the right. (I just spent 5 minutes trying to prove it was bounded because I misread the question, please kill me.) However, we have no evidence to suggest that the function's derivative is bounded to the left, so the derivative is not necessarily bounded.

For studying purposes, consider $f(x)=e^{-x}+x$, which satisfies all the properties.

The answer is easily proven to be $\boxed{C}$, from above.

Remember that bounded over a set, in this case $\mathbb R$, means that for all $x \in \mathbb R$, $|f(x)| \leq M$, for some constant M.

For example, we can firmly assert that $\displaystyle \frac{1}{x^2+1}$ is bounded over the reals, where $M=1$.

We can also say that $x$ is bounded over the set $[-5,5]$, but we cannot say that it is bounded over the reals, because $f(x)=x$ grows arbitrarily at infinity.

Answer 2

Let $\epsilon > 0$. Then there exists a natural number $M$, such that $|f'(x)-1| < \epsilon$ whenever $x>M$. Chose $M<x<y$. By the mean value theorem there exists $c\in (x,y)$ such that $$f(x)+f'(c)(y-x)=f(y)$$ Hence, $$f(y)> (1-\epsilon)(y-M)+M$$ Since you have freedom in choosing $y>M$, $f(y)$ can be made as large as possible.