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I want to gain a strong validity for formal definition of divergence compare to my version. I'll explain my current understanding below. I hope someone can help me rectifying error I made.

I was guessing out definition of divergence at a point with epsilon-delta proof (more specifically for case: $\lim_{x\to a}$ f(x)=+$\infty$) Given formal definition is $$\forall M\gt0, ( \exists\delta\gt0:0\lt\vert x-a\vert\lt\delta \Rightarrow f(x)\gt M)$$

However, the things that come up with my first trial was $$\forall\delta\gt0, (\exists\epsilon\gt0: \epsilon\lt f(x)\Rightarrow0\lt\vert x-a\vert\lt\delta)$$

Which was I think just x-y axis symmetry of convergence definition, $$lim_{x\to +\infty}f(x)=L \iff \forall\epsilon (\exists\delta\gt0: \delta\lt x\Rightarrow 0\lt\vert f(x)-L\vert\lt\epsilon)$$ I'm vaguely grasping the way to correct my fallacy such as ~ Since $lim_{x\to+\infty}f(x)=L$ is $(x\to +\infty)\to f(x)=L$, So we are not safe to say backward always works (p to q logic)~ or ~ The way how the function is defined conventionally $x\to f(x)$ so in all occasion we should argue about existence of boundary of domain not of range ~

Extra question: Is it work? $$\lim_{x\to c}f(x)=L \iff \forall \delta(\exists\epsilon\gt0:0\lt\vert f(x)-L\vert\lt\epsilon \Rightarrow 0\lt\vert x-c\vert\lt\delta)$$

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No, it does not work. Consider the function $f$ given by $f(x)=1$. Then by your definition for all $r\in\mathbb{R}$ we have $\lim_{x\rightarrow 0}f(x)=r$.

To show this let $\delta>0$ randomly given and note that for $\varepsilon=2+|r|$ we have $$|f(x)-r|\leq 1+|r|<\varepsilon$$ for all $|x|<\delta$. You will have similar problems with your proposed definition for convergence.

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  • $\begingroup$ I appreciate your disproof. It helped me to evoke extra disproof case. $\endgroup$ – WienAudience Feb 21 '19 at 13:16
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Your question is quite hard to understand but I'll try my best to answer it.

You are trying to find a definition for $ \lim_{x\to a} f(x) = + \infty$. i.e. a good definition for " if $x$ approaches $a$ then the function $f$ grows arbitrarily. The correct definition is

$$ \forall M > 0, \exists \delta > 0 : \vert x-a \vert < \delta \Rightarrow f(x) > M.$$

And you came up with $$\forall\delta\gt0, (\exists\epsilon\gt0: \epsilon\lt f(x)\Rightarrow0\lt\vert x-a\vert\lt\delta).$$

Why does this definition not work ? Think about what it's saying ! It says the following

  1. take any positive number $\delta$
  2. Consider the interval around $a$ : $(a - \delta, a + \delta)$
  3. There exists another positive number $\epsilon$ such that if $f(x)$ is greater than $\epsilon$ then $x \in (a-\delta, a+ \delta)$.

There are problems with this definition

Consider a function such that $\lim_{x \to c} f(x) = + \infty$ and $\lim_{x \to c'} f(x) = +\infty$ (in the usual sense) for example $f(x) = \frac{1}{(x+3)^2 (x-3)^2}$. According to your definition $\lim_{x \to 3} f(x) \neq +\infty$.

To see this draw the graph and notice that the function is completely symmetric so if for $\delta$ you find an $\epsilon$ such that if $f(x)$ is greater than $\epsilon$ then $x \in (a-\delta, a+ \delta)$. Then $f(x) = f(-x) > \epsilon$ so $-x \in (a- \delta, a+ \delta)$. Clearly if you take $\delta$ very small you can't have both $x \in (a-\delta, a+ \delta)$ and $-x \in (a- \delta, a+ \delta)$.

Remember that $p \Rightarrow q $ is equivalent to $ \lnot q \Rightarrow \lnot p$ so you're definition is identical to

$$ \forall \delta > 0, \exists \epsilon > 0 : x \not \in [a-\delta,a+\delta] \Rightarrow f(x) \leq \epsilon $$

Therefore any function $f$ which has $\lim_{x \to c} f(x) = + \infty$ and $\lim_{x \to c'} f(x) = +\infty$ (in the usual sense) for $c \neq c'$ will not go to infinity at all according to your definition.

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  • $\begingroup$ Thank you for your considerate answer. It truly clarified a lot in my mind. I feel i'm well convinced about theorem now. $\endgroup$ – WienAudience Feb 21 '19 at 13:13

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