I currently studying epsilon-delta proofs for proving limits of functions, and I've been struggled with those proofs.

I want to show that this claim is false:

$$ \text{if }\exists \epsilon >0,\forall \Delta >0,\forall x\in \mathbb{R} (0<\left|x-x_0\right|<\Delta \implies \left|f\left(x\right)-L\right|<\epsilon) \text{, then }\:\lim _{x\to x_0}f\left(x\right)=L$$

Can someone give me a counter example and explain it to me? I really need some guidience here.


Take $f$ the constant function equal to $1$ and $L=0$ and $\epsilon=2$ and you have a counterexample.

  • $\begingroup$ can you explain me why its counter example? $\endgroup$ – user2637293 Jan 5 '15 at 21:27
  • $\begingroup$ These values satisfy the claim but obviously the limit of this function at any $x_0$ isn't $0$. $\endgroup$ – user63181 Jan 5 '15 at 21:31
  • $\begingroup$ @RoryDaulton Because $\epsilon = 2$ so $\vert f(x)-L\rvert < \epsilon$ is true. Your expression says that there exists some $\epsilon$ such that so and so. Take $\epsilon = 2$ and see that what you wrote is satisfied. $\endgroup$ – user4894 Jan 6 '15 at 6:48
  • $\begingroup$ Yes, you are right, and I was wrong. I apologize, and I'll delete my comment in a bit. However, you may want to put your last comment as an edit to your answer. BTW, the expression is not mine, but the OP's. $\endgroup$ – Rory Daulton Jan 6 '15 at 11:45

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