# Can someone give me a counterexample to understand why this definition of limit is wrong?

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.

• Any bounded function satisfies your definition, for example. – Ian Apr 27 '15 at 1:02
• Epsilon could be very, very large.. – Cameron Williams Apr 27 '15 at 1:05

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 $(a-\delta,a+\delta)$)
• @Raquelnb Your definition does not guarantee that $f(x)$ and $f(y)$ are close when $x$ and $y$ are close, in that any bounded function, including for instance functions with jumps, satisfies it. This happens because the fact that $\varepsilon$ is existentially quantified means that it is free to be large even when $\delta$ is small. – Ian Apr 27 '15 at 1:46
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 $|x-a| < \delta$, $|f(x)-L| < \epsilon$.