🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Assume that $f$ is continuous at $a$, now we will prove that ${lim}_{h\to 0}f(a+h)=f\left(a\right)$. To show this limit exists, we refer the the definition of limit of a function. So let $ϵ\in {ℝ}^{>0}$, then since $f$ is continuous at $a$ then we know there is some ${\delta }_c\in {ℝ}^{>0}$ such that $\forall x\in dom\left(f\right),|x-a|<{\delta }_c⟹|f\left(x\right)-f\left(a\right)|<ϵ$.

We need to prove that there is some $\delta \in {ℝ}^{>0}$ such that $\forall h\in dom\left(f\right),|h-0|<\delta ⟹|f(a+h)-f\left(a\right)|<ϵ$, so suppose that $h\in \text{dom}\left(f\right)$, thus we can also consider the value of $h+a\in \text{dom}\left(f\right)$. Thus by the assumption of continuity, as it holds for all $x$ in $\text{dom}\left(f\right)$ then $|(h+a)-a|<{\delta }_c⟹|f(a+h)-f\left(a\right)|<ϵ$, therefore taking $\delta ={\delta }_c$ will complete this direction of the proof.

Now we work in the other direction, first assuming that ${lim}_{h\to 0}f(a+h)=f\left(a\right)$ and trying to show that $f$ is continuous at $a$, so let ${ϵ}_c\in {ℝ}^{>0}$, thus since the aformentioned limit exists, then we have some $\delta \in {ℝ}^{>0}$ such that $\forall h\in \text{dom}\left(f\right),|h-0|<\delta ⟹|f(h+a)-f\left(a\right)|<{ϵ}_c$.

Working in s similar fashion as in our first direction, we consider the value $x-a\in \text{dom}\left(f\right)$, since the latter statement in the previous paragraph is for all $h\in \text{dom}\left(f\right)$, we can consider $h=x-a$, which says that $|x-a-0|<\delta ⟹|f(x-a+a)-f\left(a\right)|<{ϵ}_c$ which is equivalent to $|x-a|<\delta ⟹|f\left(x\right)-f\left(a\right)|<{ϵ}_c$, which is exactly what we wanted to prove.