# Limits and continuity: a real analysis proof.

Let $E \subseteq R$, $p$ is limit point of $E$ , and $f\colon E\to R$. Suppose there exist a constant $M>0$ and $L\subseteq R$ such that $|f(x)-L| \leq M|x-p|$ for all $x\subseteq E$. Prove $\lim_{x\to p} f(x) = L$.

My approach is as follows:

Proof:

Since $p$ is a limit point of $E$ and $E\subset R$. $x \to p$ so $0 < |x - p| < \epsilon$ and $|f(x) -L|\leq M|x-p|$ $|f(x)-L| \leq M \epsilon$ Then suppose there exists a constant $M > 0$ , if we choose $M=1$ then $|f(x)-L| \leq \epsilon$ so $\lim_{x\rightarrow p} f(x) = L$ ?

I don't think its right, can someone help me on how else I should approach this proof?

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Welcome to math.SE! Here's a quick guide on how to write math on this site: meta.math.stackexchange.com/questions/5020/… –  Newb Nov 9 '13 at 21:07

Since $p$ is a limit point of $E$, there exist a sequence $\{x_n\}\in E$ such that $\{x_n\}\rightarrow p$. Then, given any $\epsilon>0$, there exist an $N\in\mathbb{N}$ such that $|x_n - p|<\frac{\epsilon}{M}$ for $n>N$. From your assumption about $f$, $|f(x_n)-L|\leq M|x_n -p|$ for all $x_n \in E$. Then, by definition of limit $\lim_{x\rightarrow p}f(x) = \lim_{n\rightarrow \infty} f(y_n)$ whenever $y_n \rightarrow p$. So, consider a sequence $x_n\rightarrow p$, then we have
$|f(x_n)-L|\leq M|x_n-p| < \epsilon$,
when $n>N$ for the correct choice of $N$, but we are guaranteed that one exists! This holds for any choice of $\epsilon$, so $f(x_n)\rightarrow L$, or $\lim_{n\rightarrow\infty}f(x_n)\rightarrow L$, as desired. So you were very close.