Proof Clarification: $\lvert f(x)-L\rvert$ $\le$ K$\lvert x-c\rvert$ implies $\lim \limits_{x \to c}$ f(x) = L. Let A be an interval in the real numbers, let f: A -> $\mathbb{R}$ and let c $\in$ A. Suppose there exist constants K and L such that $\lvert f(x)-L\rvert$ $\le$ K$\lvert x-c\rvert$ for each x $\in$ A. Show that $\lim \limits_{x \to c}$ f(x) = L.
Proof:
Since A $\subseteq$ $\mathbb{R}$, we know that any c $\in$ A is a limit point of A. Suppose that for all $\epsilon$ $\gt$ 0, there is a $\delta$ $\gt$ 0 such that 0 $\lt$ $\lvert x-c\rvert$ $\lt$ $\delta$ for all x $\in$ A.
Since 0 $\lt$ $\lvert x-c\rvert$ $\lt$ $\delta$, then 0 $\lt$ K$\lvert x-c\rvert$ $\lt$ K$\delta$. Since K$\delta$ $\gt$ 0 and $\delta$ is arbitrary, we have $\lvert f(x)-L\rvert$ $\lt$ $\delta$ so that $\lim \limits_{x \to c}$ f(x) = L by defintion. $\blacksquare$

My question is whether what I have bolded above is sufficient for all $\epsilon$ $\gt$ 0 to imply the rest of my argument. If it's not, then how do I proceed?
 A: It’s not quite right. Let’s see, let $c \in A$, you need to prove that given $\varepsilon>0$, there exists $\delta>0$ such that $$0<|x-c|<\delta \implies |f(x)-L|<\varepsilon.$$ If we take $\delta:=\frac{\varepsilon}{K}$ then $$|f(x)-L|\leq K|x-c|<K \cdot \frac{\varepsilon}{K}=\varepsilon.$$ This proves that $\displaystyle \lim_{x \to c} f(x)=L$.
A: There are a few problems in your solution.


*

*you wrote $\lim\limits_{x\to x} f(x)$, it should be $x\to c$. 


*The statement of first line is not clear, we do not know if it means 
$\exists (K,L)\mid \forall x\in A, |f(x)-L|<K|x-c|$ supposedly correct version or $\forall x\in A,\exists (K,L) \mid |f(x)-L|<K|x-c|$ in which case $K$ and $L$ depends of $x$.


*The second paragraph is quite weird. We do not suppose anything, we select an arbitrary $\varepsilon$ and impose $|x-c|<\delta$.


*In the last paragraph you lost trace of $\varepsilon$, where has it gone?! Remember that you are supposed to upper bound $|f(x)-L|$ with this $\varepsilon$.
So it should go like this.
Let $A$ be an interval of $\mathbb R$ and $f: A\mapsto \mathbb R$
For $c\in A,L\in\mathbb R,\exists K>0\mid \forall x\in A,|f(x)-L|<K|x-c|$
Then $\forall \varepsilon>0$, let's choose $\delta=\dfrac{\varepsilon}K$ then  $|x-c|<\delta\implies |f(x)-L|<K|x-c|<K\delta=\varepsilon$
So $\lim\limits_{x\to c} f(x)=L$.


*

*Last issue, is what above is only well defined for $c\in\mathring A$, and if $\inf A=\alpha$ and $\sup A=\beta$ then we also have to choose $\delta<\min(c-\alpha,\beta-c)$ for the $K$ inequality to stay true. 


If $A$ is a closed interval, then we would only have $\lim\limits_{x\to \alpha^+} f(x)=L$ and $\lim\limits_{x\to \beta^-} f(x)=L$ for the boundary points of $A$.
