Is there a function such that $f:\mathbb{R}\rightarrow\mathbb{R}$ and $\lim_{x\rightarrow2}f(x)=4$ but it doesn't satisfy that: $(\exists\delta>0)(\forall\epsilon>0)(\forall x\in\mathbb{R})(0<\vert x-2\vert<\delta)\Rightarrow(\vert f(x)-4\vert<\epsilon)$

I tried to look at the contrary which is: $(\forall\delta>0)(\exists\epsilon>0)(\exists x\in\mathbb{R})(0<\vert x-2\vert<\delta)\wedge(\vert f(x)-4\vert\ge\epsilon)$

Which I guess I need to find a bounded function. I'd like to get an example to help me with my intuition about the this.

Thanks in advance!

  • $\begingroup$ Didn't you mean $(\forall\epsilon>0)(\exists\delta>0)$ in the first line? $\endgroup$ – The Phenotype Jan 4 '18 at 2:28
  • $\begingroup$ Nope. but I added that the limit is 4 when $x \rightarrow 2$. Please take a look now $\endgroup$ – user401575 Jan 4 '18 at 2:29

Yes - let $f(x) = 2x$. For any $\delta$, choose $x = 2 + \frac{\delta}{2}$ and $\epsilon = \frac{\delta}{2}$


This is true for any continuously differentiable function at a point $a$ with $f'(a) \ne 0$.

Let $f(a) = b$ and $f'(a) = c \ne 0$.

Then, for small $h$, $\dfrac{f(a+h)-f(a)}{h} \approx f'(a) = c$ so $f(a+h)-f(a) \approx hf'(a) $.

Your expression becomes $(\exists\delta>0)(\forall\epsilon>0)(\forall x\in\mathbb{R})(0<\vert x-a\vert<\delta)\Rightarrow(\vert f(x)-f(a)\vert<\epsilon) $


$(\exists\delta>0)(\forall\epsilon>0)(0<h<\delta)\Rightarrow(\vert hf'(a)\vert<\epsilon) $.

However, the condition states $(\forall\epsilon>0)$. Since $f'(a) \ne 0$, for any $h > 0$ we can choose $\epsilon < |hf'(a)|$ so the expression is false.

This is a little sloppy, but it can be made more rigorous by making $f$ have a continuous derivative and by choosing a neighborhood of $a$ such that

$0 \le |\dfrac{f(a+h)-f(a)}{h}-f'(a)| \le \frac12|f'(a)| $ so $|f(a+h)-f(a)-hf'(a)| \le \frac12|hf'(a)| $.

If we then choose $\epsilon \lt \frac12|hf'(a)| $ we again get a false statement.


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