Epsilon-Delta definition

Question

The defintion give me a confusion. It states that for all $\epsilon>0$ there exists a $\delta>0$ such that, whenever $|x−a|<\delta$ then $|f(x)−L|<ϵ$. So when I think about $|x−2|<\delta$ $\to$ $|x^2 -1| < \epsilon$. It is also true for every $\epsilon >0$ since I just need to choose large $\delta$ such that $f(x)=x^2$ on $|x−2|<\delta$ contain $|x^2-1|<\epsilon$

Answer 1

The limit in the example is $$\lim_{x \to 2} x^2 = 1.$$ The definition of the limit is: $$\forall \epsilon> 0, \exists \delta > 0, |x-2|<\delta \Rightarrow |x^2-1|<\epsilon.$$

First of all, when we see $0<|x-2|<\delta$, we talk about $x$-coordinate. But when we see $|x^2-1|<\epsilon$, we talk about $y$-coordinate.

But, I will follow your thought. Look at the quantifier $\forall \epsilon$. If you choose large $\delta$, for instance $\delta = 10$. So that $|x-2|<10 \Longleftrightarrow -8<x<12$. We may choose $x = 8$ so, $$|x^2-1| = |8^2-1|=63.$$ Because the quantifier of $\epsilon$ is $\forall$ positive epsilon, we may choose $\epsilon = 1$. It is clear that $$|x^2-1| = 63 > \epsilon.$$

So the definition of limit is not true. In fact, the limit is not equal to 1.