Krdytkyu

Let $I$ be an interval in $mathbb{R}$ such that $f:Irightarrowmathbb{R}$ is a function and let $cin I.$
Show that $|f(x)-L|leq K|x-c| implieslim_{xto c} f(x)=L$

Proof:

Suppose w.l.o.g. that $K>0$. Then we have $K|x-c|>0$ since $|x-c|>0$. Now since $xin I$ and $cin I$, $existsepsilon>0$ 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 $lim_{xto c}f(x)=L$. $blacksquare$

Is this proof valid?

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 $existsvarepsilon>0,|x-c|<dfrac{varepsilon}{K}$. 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 $existsvarepsilon$, so $varepsilon$ is not arbitrary, but rather specifically chosen.

Here's what you want to prove:

$$forallvarepsilon>0,existsdelta>0,forall xin I,|x-c|<deltaimplies |f(x)-L|<varepsilon.$$

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:

$$existsdelta>0,forall xin I,|x-c|<deltaimplies |f(x)-L|<varepsilon.$$

And here, your choice of $delta=dfrac{varepsilon}{K}$ does the work.

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
$$|f(x) - L| leq K |x - c| < Kcdot epsilon.
$$
Now let us set $tilde{epsilon} := Kcdot epsilon > 0$, then for any $x$ close enough to $c$ such that $| x - c| < epsilon$, this implies that $|f(x) - L| < tilde{epsilon}$.