I want to disprove : $$\lim_{x\to 0^+}\frac{1}{x} = c \in R$$ using epsilon-delta definition. So here is what I've done :
I know that $c>0$ ( because $\frac{1}{x}>0$ if $x>0$ ).
So applying $\epsilon ,\delta $ definition to check whether the limit is verified , namely : $$\forall\epsilon>0\,\exists\,\delta >0 \,\backslash\,0<x<\delta\Rightarrow-\epsilon<\frac{1}{x}-c<\epsilon$$
So taking $0<\epsilon<c$ :
$$-\epsilon<\frac{1}{x}-c<\epsilon \Rightarrow \frac{1}{c-\epsilon} < x < \frac{1}{c+\epsilon}\;$$ which is a contradiction of $x \rightarrow0^+$ if $\epsilon\rightarrow 0$ . So this limit can't be finite.
Is this reasoning correct? How can I write it in a more formal way that the last inequality implies a contradiction?
Best Answer
Your reasoning is correct. I would fix up the language to be more precise and conventional.
You want to show that it is not true that there exists $c \in \mathbb{R}$ such that for all $\epsilon > 0$ there exists $\delta > 0$ such that for all $x$, $0 < x < \delta \implies \left|\frac{1}{x}-c\right| < \epsilon$.
When you negate a quantified statement, the quantifiers get flipped. So you want to show: for all $c \in \mathbb{R}$, there exists $\epsilon > 0$ such that for all $\delta > 0$, there exists $x$, such that $ 0 < x < \delta$ and $\left|\frac{1}{x}-c\right| \geq \epsilon$.
You wrote “taking $0 < \epsilon < c$,” but I would recommend being specific. You are trying to satisfy a “there exists” quantifier, so point one out. $\epsilon = \frac{c}{2}$ works pretty well.
Now, given $\delta > 0$, you want to find $x$ such that $0 < x < \delta$ and $\left|\frac{1}{x}-c\right| \geq \frac{c}{2}$. As long as $0 < x < \frac{1}{2c}$, then $\left|\frac{1}{x}-c\right| > c > \frac{c}{2}$. So let $x$ be the minimum of $\frac{1}{2c}$ and $\frac{\delta}{2}$. That way both $0 < x < \delta$ and $0 < x < \frac{1}{2c}$ are satisfied.
This takes care of the case that $c > 0$. What about $c = 0$? (You asserted that the limit must be positive; that's not true.) $c < 0$?