real analysis - For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$

I saw the following claim in this thread :

For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.

A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$,$b$ and $\lambda \in [f(a),f(b)]$ there is a $x \in [a,b]$ such that $f(x)=\lambda.$

A failed attempt:

Let $f$ be a real function on $\mathbb{R}$ with the intermediate property. Let $p$ be a point where the function possesses left hand and right hand limits $f(p)_-, f(p)_+.$

To begin with I should be able to show that $f(p)_- = f(p)_+=l_p$(which I am not able to show)

Once that is shown, I tried this . . .

let $(x_n)$ be a sequence in $(-\infty,p)$ converging to $p$ and $(y_n)$ be a sequence in $(p,\infty)$ converging to $p.$

Then $x_n < y_n$ for each $n\in \mathbb{N}.$

For each $n$ let $\lambda_n\in \mathbb{R}$ be such that $\lambda_n \in [f(x_n),f(y_n)].$ By the intermediate value property there exists $p_n \in (x_n,y_n)$ such that $f(p_n)=\lambda_n.$

It is evident using sandwich theorem that $p_n$ converges to $p$ and $\lambda_n=f(p_n)$ converges to $l_p$

. . . and this is leading nowhere

Can anyone show me a way to do this?

Answer

Assume $f(p)_- \neq f(p)_+$, wlog $f(p)_- < f(p)_+$. Let $c:=f(p)_+ - f(p)_-$

Then by definition of the one-sided limit there is $\varepsilon > 0$ such that

$$f(x)and
$$f(x) > f(p)_+ - \frac{c}{4} \, \text{ if } x\in (p, p+\varepsilon)$$

It should now be easy to verify that this will imply a contradiction to the intermediate value property.