How to show that $x, y$ $\in$ $\mathbb{R}$ exist such that $f(x) < 0 < f(y)$ using a Epsilon-Delta proof?

Question

For a small part of a school assignment (real Analysis) I have to show that $x, y$ $\in$ $\mathbb{R}$ exist such that $f(x) < 0 < f(y)$

Let $f$: $\mathbb{R}$ $\longrightarrow$ $\mathbb{R}$ differentiable in $0$ and assume that $f(0)$ = $0$ and $f'(0) = 1$

I already have something like this:

$$\lim\limits_{x\to0}\frac{f(x)-f(0)}{x-0} = f'(0)2$$ in other words:

$$\lim\limits_{x\to0}\frac{f(x)-0}{x} = \lim\limits_{x\to0}\frac{f(x)}{x} = 1$$

en then I would think: $$\left| \frac{f(x)}{x} - 1 \right|....$$

and then I need to prove this with an Epsilon-Delta proof, but I got stuck.

Answer 1

By definition of limit, with $\varepsilon = \frac12$, there is a $\delta>0$ such that $$ -\frac12<\frac{f(x)}x - 1< \frac12\\ \frac12 < \frac{f(x)}x < \frac32 $$ for all $x$ with $0<|x|<\delta$. Among these $x$, pick one that is negative and one that is positive.

Answer 2

Let $\epsilon>0$ with $\epsilon<1$.

By definition of the limit you showed existing and converging to $1$, there exists $\delta>0$ such that if $0<|x|<\delta$, then $$ 1-\epsilon<\frac{f(x)}{x}<1+\epsilon. $$ When $x$ is positive ... and when $x$ is negative ...