applying epsilon delta Let $I \subset \mathbb R$ be an open domain, $x_0 \in I$ and $f: I \to \mathbb R$ function. Let's suppose that there exists constants $K > 0$ and $L \in \mathbb R$ so that 
$|f(x)-L| \le K|x-x_0|$ for all $x \in I$.
Show with epsilon/delta that $\lim_{x \to x_0} f(x)=L$
I know this is a straight apply to definition of epsilon/delta, but maybe someone could show me how to do it?
 A: Usually to do these $\epsilon$-$\delta$ proofs, you let $\epsilon > 0$ be arbitrary and write down what you want to show:
We want to find some $\delta>0$ such that if $|x-x_{0}|< \delta$, then $|f(x) - f(x_{0})|< \epsilon$.
So we want $|f(x) - f(x_{0})|< \epsilon$.  What are we given in the problem?  $\underbrace{|f(x) - f(x_{0})|}_{\text{want this } < \epsilon} < \underbrace{K|x-x_{0}|}_{\text{Need to choose } \delta \,\text{to get this } \leq \epsilon}$.
Okay, well a natural thing to do then is to choose $\delta$ to be $\frac{\epsilon}{K}$.  Then if $|x- x_{0}| < \delta$, we have:
$|f(x) - f(x_{0})| < K|x-x_{0}| < K\frac{\epsilon}{K} = \epsilon$.  So we found that if for each $\epsilon > 0$, we take $\delta = \frac{\epsilon}{K}$, then $|x- x_{0}| < \delta$ implies $|f(x) - f(x_{0})| < \epsilon$.
A: First, note that since $I$ is open, we know that there exists some $R > 0$ such that:
$$
(x_0 - R, x_0 + R) \subseteq I
$$
Now given any $\epsilon > 0$, let $\delta = \min\{R, \epsilon/K\}$, which is certainly positive. Then observe that if $|x - x_0| < \delta$, then:
\begin{align*}
|f(x) - L|
&\leq K|x - x_0| &\text{since $\delta \leq R$, so $(x_0 - \delta, x_0 + \delta) \subseteq (x_0 - R, x_0 + R) \subseteq I$} \\
&< K(\epsilon/K) &\text{since $|x - x_0| < \delta \leq \epsilon/K$} \\
&= \epsilon
\end{align*}
as desired. $~~\blacksquare$
