Could someone give me a counterexample to understand why this definition of $\lim_ {x\to a} f(x) = L$, does not work?
$\forall \delta>0 \exists \varepsilon>0$ such that, if $0<|x−a|<\delta$, then $|f(x)−L|<\varepsilon$
I've searched in this forum and the question has already been made but I don't understand the examples and counterexamples given. I found this: $\lim_{x\to 1} \frac{1}{x} = 1$, as an example in Spivak Supplement of calculus, but I don´t understand why this example works.
Suppose $f$ is bounded, $f(x)\leq M$ for every $x$. Then, following your definition, $\lim_{x\to a}f(x)=A$ for every $a$ and every $A$. You give me a $\delta$. I choose $\varepsilon = M+|A|$. Then, if $0<|x-a|<\delta \to |f(x)-A|\leq|f(x)|+|A|<M+|A|=\varepsilon$.
(Note that the $\delta$ you chose doesn't even matter, given that $|f(x)-A|\leq|f(x)|+|A|<M+|A|$ holds for every $x$, in particular to those in $(a-\delta,a+\delta)$)
You want to show that you can get $f(x)$ as close as you want to $L$ by choosing $x$ close enough to $a$.
This means that, using your notation, for any $\epsilon > 0$ you can choose a $\delta$, which is usually a function of $\epsilon$, such that whenever $|x-a| < \delta$, $|f(x)-L| < \epsilon$.
In your logical notation, this is ∀ ϵ > 0 ∃ δ>0 . Translation: For all ϵ > 0 there is a δ>0 such that whenever $|x-a| < \delta$, $|f(x)-L| < \epsilon$.
