Show that $|f(x)-L|\leq K|x-c| \implies\lim_{x\to c} f(x)=L$ Let $I$ be an interval in $\mathbb{R}$ such that $f:I\rightarrow\mathbb{R}$ is a function and let $c\in I.$
Show that $|f(x)-L|\leq K|x-c| \implies\lim_{x\to c} f(x)=L$
Proof:
Suppose w.l.o.g. that $K>0$. Then we have $K|x-c|>0$ since $|x-c|>0$. Now since $x\in I$ and $c\in I$, $\exists\epsilon>0$ S.t. $|x-c|<\frac{\epsilon}{K}$; we can make the distance between $x$ and $c$ as small as we like. Therefore by assumption, $$|f(x)-L|<K|x-c|<K(\frac{\epsilon}{K})=\epsilon$$
Whence, we conclude since $\epsilon$ was arbitrary, that $\lim_{x\to c}f(x)=L$. $\blacksquare$
Is this proof valid? 
 A: You have the right ideas, well done! But unfortunately there are some points in the proof that are wrong.
You say that $K|x-c|>0$ because $|x-c|>0$. But if you take $x=c$, this is no longer true.
You also say that $\exists\varepsilon>0,|x-c|<\dfrac{\varepsilon}{K}$. This statement is true, but it doesn't mean that $\varepsilon$ can be as small as you want unlike what you said. In fact, such an $\varepsilon$ has to satisfy $K|x-c|<\varepsilon$, so epsilon can't be as small as, say, $\frac{1}{2}K|x-c|$. The statement also is of the form $\exists\varepsilon$, so $\varepsilon$ is not arbitrary, but rather specifically chosen.
Here's what you want to prove:
$$\forall\varepsilon>0,\exists\delta>0,\forall x\in I,|x-c|<\delta\implies |f(x)-L|<\varepsilon.$$
So the natural way to start your proof is: Let $\varepsilon>0$. Now that's a truly arbitrary. So what you need to prove is this:
$$\exists\delta>0,\forall x\in I,|x-c|<\delta\implies |f(x)-L|<\varepsilon.$$
And here, your choice of $\delta=\dfrac{\varepsilon}{K}$ does the work.
A: Yes it is valid once you take into account Sean Roberson's comment that $K>0$. Then for $f:I \rightarrow \mathbb{R}$ such that for some $c \in I$ and some $L \in \mathbb{R}$, we have that $|f(x) - L| \leq K |x - c|$. For any $\epsilon > 0$, choose any $x \in I$ close enough to $c$ such that $|x - c| < \epsilon$. Then
$$|f(x) - L| \leq K |x - c| < K\cdot \epsilon.
$$
Now let us set $\tilde{\epsilon} := K\cdot \epsilon > 0$, then for any $x$ close enough to $c$ such that $| x - c| < \epsilon$, this implies that $|f(x) - L| < \tilde{\epsilon}$.
