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.