Result regarding signs and a limit

[Mr Davis 97]

Suppose that $f : \mathbb{R} \to \mathbb{R}$ is differentiable at $a\in\mathbb{R}$. Is it true that if $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}>0$ and $x>a$ then $f(x)>f(a)$? I'm trying to find a counterexample to show that its false because I think it is, but I'm having a hard tome doing that for some reason. The reason I am asking is because it was used in someone's proof with no justification, but I think it's wrong.

 

[fresh_42]

Suppose that $f : \mathbb{R} \to \mathbb{R}$ is differentiable at $a\in\mathbb{R}$. Is it true that if $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}>0$ and $x>a$ then $f(x)>f(a)$? I'm trying to find a counterexample to show that its false because I think it is, but I'm having a hard tome doing that for some reason. The reason I am asking is because it was used in someone's proof with no justification, but I think it's wrong.

Why do you think it's wrong? Say we have $L := \lim_{x\to a}\frac{f(x)-f(a)}{x-a}>0\,.$ Then for sufficiently close values $x$ to $a$, we have $0 < L - \frac{L}{3} < \frac{f(x)-f(a)}{x-a} < L + \frac{L}{3}$.

 

[Mr Davis 97]

Why do you think it's wrong? Say we have $L := \lim_{x\to a}\frac{f(x)-f(a)}{x-a}>0\,.$ Then for sufficiently close values $x$ to $a$, we have $0 < L - \frac{L}{3} < \frac{f(x)-f(a)}{x-a} < L + \frac{L}{3}$.

Okay, I think I see. But is this result obvious enough just to claim that since $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}>0$ and $x>a$ implies $f(x)>f(a)$ without further comment on why exactly this implication is true? Is it something that should be proved in a lemma or is it just an obvious result?

 

[fresh_42]

Okay, I think I see. But is this result obvious enough just to claim that since $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}>0$ and $x>a$ implies $f(x)>f(a)$ without further comment on why exactly this implication is true? Is it something that should be proved in a lemma or is it just an obvious result?

I found it obvious as I thought about it this way: In order for the nominator to be negative, the entire quotient would have to be negative. But as we approach a strictly positive number, we will sooner or later have to leave the range of negativity: we cannot wait until the very last moment and all of a sudden jump onto a positive limit - at least not with a continuous function, which it is, as it's differentiable there.