I found the following problem a little tricky and I'd be glad if I could get some direction (just a hint):

Let f be a function defined around $x_o$. For every $\epsilon>0$ there's some $\delta>0$ such that if $0<|x-x_0|<\delta$ and $0<|y-x_0|<\delta$ then $|f(x)-f(y)|<\epsilon$.

What's needed to be proven is that $\lim_{x\to x_0}f(x)$ exists.

A big thanks in advance.

P.S. I've looked at the opposite question, where you need to prove what is given here, and you are given what is needed to prove here. It was easier as you could just add $L$ & $-L$ inside the absolute value and then just use the triangle inequality and play with the epsilon, but proving the opposite (what I asked above) appears to be trickier.

  • $\begingroup$ I guess that you are talking about $\lim_{x\to x_0} f(x)$. Do you know about triangular inequality? $\endgroup$
    – Sigur
    Nov 10 '12 at 21:53
  • $\begingroup$ Oops, I've now fixed the question. Also, I did try to use the triangle inequality, but without interesting results. $\endgroup$
    – Py42
    Nov 10 '12 at 21:58
  • $\begingroup$ @Sigur: note that the property asked for does not give you the limit, so you cannot do any inequalities with it: you first have to prove it exists. $\endgroup$ Nov 10 '12 at 23:30
  • $\begingroup$ @MartinArgerami: I think he meant using the inequality in order to prove it exists (although I don't know how it can be done). $\endgroup$
    – Py42
    Nov 10 '12 at 23:48

Consider the sequence $f(x(0)+1/2),f(x(0)+1/3),f(x(0)+1/4),....$. From the conditions of the problem, it follows that this is a cauchy sequence and thus converges. Let the previous sequence converges to L. Now use the triangle inequality to deduce that $f(x)$ tends to L as x approaches a.

  • $\begingroup$ Thank you. Do you happen to know if there is also a non-sequence approach? $\endgroup$
    – Py42
    Nov 10 '12 at 22:05

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