Prove that a Lipschitz function is continuous Define $f:[a.b] \rightarrow \mathbb{R}$ as Lipschitz
By definition, then, $\exists M&gt;0, M \in \mathbb{R}:|f(x)-f(y)\leq M|x-y|$  $ $ $\forall x,y\in[a,b]$
Rearranging this, we have $\frac{|f(x)-f(y)|}{x-y} \leq M$, positing that the slope must be finite on the interval $[a,b]$
I understand conceptually that Lipschitz functions must be continuous but I'm having trouble showing it.
Should I go about this by showing assuming the negation of Lipschitz and showing that for some $\epsilon,  \exists \delta&gt;0: |f(x)-L|&gt;\epsilon$ ?
It is not necessary to proceed by contradiction. Use the $\varepsilon-\delta$ definition of continuity and choose $\delta=\varepsilon/M$.
If $\delta=\frac{\epsilon}{M}$. The definition of continuity says that we can suppose that $$0&lt;|x-x_0|&lt;\delta$$ from here: $$M|x-x_0|&lt;\epsilon$$ but we now that $$|f(x)-f(x_0)|&lt;M|x-x_0|$$ because $f\in Lip(\mathbb{R})$, then $|f(x)-f(x_0)|&lt;\epsilon$ and therefore $f$ is continuous.