Prove corollary of Darboux's Theorem I'm currently going through some analysis problems in preparation for an upcoming test, and I was struggling with the following problem:
Let ${f:[a,b]} \rightarrow \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ and let $x \in (a,b)$. Suppose $lim_{t \rightarrow x^{-}}f'(t)=L_{1}$ and $lim_{t \rightarrow x^{+}}f'(t)=L_{2}$.
Prove $L_1=L_2.$
Now, I can see that this likely follows from a contradiction, if we can apply Darboux's Theorem after assuming $L_1<L_2$ (without loss of generality), but I'm a little confused by how to do such. Any help is appreciated!
 A: Prove by contradiction. Suppose the contrary that $L_{1}\neq L_{2}$.
Consider the case that $L_{1}<L_{2}.$ (The case $L_1>L_2$ can be treated similarly.) Choose $c,d\in\mathbb{R}$
such that $L_{1}<c<d<L_{2}$. By property of limit, there exists $\delta>0$
such that $f'(t)<c$ whenever $t\in(x-\delta,x)$ and $f'(t)>d$ whenever
$t\in(x,x+\delta)$. Choose $x_{1}\in(x-\delta,x)$ and $x_{2}\in(x,x+\delta)$.
Choose $y\in(c,d)\setminus\{f'(x)\}$. Observe that $f'(x_{1})<y<f'(x_{2})$,
so, by Darboux's Theorem applied on $[x_1,x_2]$, there exists $\xi\in(x_{1},x_{2})$ such
that $f'(\xi)=y.$ Clearly $\xi\neq x$ because $y\neq f'(x)$. If
$\xi\in(x_{1},x),$ then $f'(\xi)<c$. If $\xi\in(x,x_{2})$, then
$f'(\xi)>d$. Either case contradicts to $f'(\xi)=y$.
A: Suppose $f'(x)=L$. Find $m\in \mathbb{N}$ so that $x+\frac{1}{n}\in (a,b)$ for all $n\geq m$. From MVT we have the existence of $c_n\in \Big(x,x+\frac{1}{n}\Big)$ such that $$n\Bigg(f\Big(x+\frac{1}{n}\Big)-f(x)\Bigg)=f'(c_n)$$ Since $\lim_{t \rightarrow x^{+}}f'(t)=L_2$ and $c_n \rightarrow x^+$ we have $f'(c_n)\rightarrow L_2$. Moreover, since we know that $f'(x)$ exists and equals $L$, we have $n\Bigg(f\Big(x+\frac{1}{n}\Big)-f(x)\Bigg) \rightarrow L$ telling us that $L=L_2$. An identical argument considering a sequence of the form $x-\frac{1}{n}$ will tell us $L=L_1$.
