# Epsilon-delta condition for defining divergence

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

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.

• I appreciate your disproof. It helped me to evoke extra disproof case. – WienAudience Feb 21 '19 at 13:16

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.

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