Given a continous function $f$ Prove that $f(I)=\mathbb R$ . Let $f$ be a continuous function on $I=(0,1)$ such that $$\lim_{x\to0^+}f(x)=-\infty\text{ and }\lim_{x\to1^-}f(x)=\infty$$
Prove that $f(I)=\mathbb R$.
To me this intuitive, but we should prove it rigoursly
The limit conditions are equivalent to $\forall M_1>0, \exists\delta_1>0$ such that $$0<x<\delta_1 \implies f(x)<-M_1$$
and $\forall M_2>0, \exists\delta_2>0$ such that $$0<1-x<\delta_2 \implies f(x)>M_2$$
But I can't see where I can use that?
 A: All you need to show is that an arbitrary value $r\in \mathbb{R}$ has a preimage in $(0,1)$.    Assume it doesn't, use that to show a contradiction in one of your two limits along with the intermediate value theorem.]
A few more details:   By the intermediate value theorem,  if there exists a value on either side of $r$ we are done,  so assume that all the values are less than $r$.  This contradicts the limit going to $\infty$.   Similar for the other side
A: The strategic point to remember is, with the definitions in question we "can make any workable choice smaller without breaking it."
Here is a direct proof sketch in the notation of the question, broken out into idiomatic steps whose proofs you're invited to supply.

*

*For every real number $y$, there exists a positive real number $M$ such that $-M < y < M$.

Hint 1: Ensure $|y| < M$.

 For example, let $M = 1 + |y|$. Then $-M < -|y| \leq y \leq |y| < M$.


*

*There exists an interval $[a, b] \subset (0, 1)$ such that $f(a) < y < f(b)$.

Hint 2a:

 Ensure $f(a) < -M$ and $M < f(b)$.

Hint 2b:

 Put $M_{1} = M_{2} = M$, pick positive reals $\delta_{1}$ and $\delta_{2}$ as in the question, and also satisfying $\delta_{1} < 1 - \delta_{2}$. (Why can we assume the last inequality without loss of gnerality?) Define $a = \frac{1}{2}\delta_{1}$ and $b = 1 - \frac{1}{2}\delta_{2}$. (Why does this choice work?)


*

*There exists a real $x$ in $(0, 1)$ such that $f(x) = y$.

Hint 3: If necessary, re-read the comments. ;)

 Apply the intermediate value theorem to $f$ on the closed, bounded interval $[a, b]$.

