# 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 $\lim_{x\to p} f(x) = L$.

I'm not sure how to start with this proof. Do I prove it by contradiction? Does it follow by definition?

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

## 1 Answer

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.

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thank you this is very clear – james Miler Nov 10 '13 at 4:50
@sayuri Glad to help. I see that you have yet to accept an answer on SE. If someone answers your question to your satisfaction it is good practice to accept the answer to show others that you are pleased with the question. This also lets others know that may be asking the same question that that method worked. – Jeremy Upsal Nov 10 '13 at 15:50