Question: Let $I \subseteq \mathbb{R}$ be an open interval e that $a \in I$. Assume that $f:I\to \mathbb{R}$ is a function. Suppose that constants $K > 0$ and $L \in \mathbb{R}$ exist, such that $|f(x)-L| \leq K|x-a|$, for every $x \in I$. Show that $\lim\limits\_{x \to a} f(x) = L.$

I've been trying to prove this, but wasn't able to find a way to start this. I've tried researching, (apologies in advance if this turns out to be duplicate) but due to the many parts this question has, I haven't been able to find anything that would give me a headstart.

Could anyone give any help, or a hint on how I may start to tackle this?

Answer: 
**HINT**

What is the definition of limit? Here it is:

\begin{align\*}
\lim\_{x\to a}f(x) = L \Longleftrightarrow (\forall\varepsilon > 0)(\exists\delta\_{\varepsilon} > 0)(\forall x\in I)(0 < |x - a| < \delta\_{\varepsilon} \Rightarrow |f(x) - L| < \varepsilon)
\end{align\*}

where $a$ is an accumulation point of $I$.

Based on the given assumption, it suffices to take $\delta\_{\varepsilon} = \varepsilon/K$.

Can you take it from here?
Think about it this way, |f(x)-L| is bounded by K|x-a|, so take the limit as x approaches a.
This is $\lim\_{x\rightarrow a} |f(x)-L| \leq \lim\_{x\rightarrow a}K|x-a|$, now you got to evalute limits and remember that the absolute value function is continuous.