I wanted to double check my understanding of a statement written with delta-epsilon. Here is the question:
Question:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and let $a \in \mathbb{R}$.
Suppose
$$ \forall \delta>0, \exists b > 0, \exists x_1,x_2 \in \mathbb{R}, 0 < | x_1 – a | < \delta, 0 < | x_2 – a | < \delta, | f(x_1) – f(x_2)| \geq b$$
Does $\lim_{x \to a} f(x)$ exist?
My attempt:
It appears that no matter how small $\delta$ is, the function values around $a$ must not be constant, so that we can find a small $b$ and have $f(x_1)$ and $f(x_2)$ differ greater than this small $b$.
But I can come up with two functions that satisfy this condition.
For example, if $f(x) = x$, the delta-epsilon statement above would be true. I can just choose $x_1,x_2$ such that
$x_1 > x_2$ and $0 < | x_1 – a | < \delta$ and $0 < | x_2 – a | < \delta$. Choose $b = \frac{x_1-x_2}{2}$. In this case, the limit exists, $\lim_{x \to a} x = a$.
On the other hand, if I had a piecewise function: $f(x) = 1$ when $x \geq a$, $f(x) = -1$ when $x < a$. I would just choose $b=1$, $x_1 \geq a$ and $x_2 < a$, this also satisfies the delta-epsilon statement above. In this case, the limit does not exist, since the left hand and right hand limits are different as $x$ approaches $a$.
Is this correct?
Best Answer
Yes, the epsilon-delta condition simply says that $f$ is not constant in any disc about $a$.
Then the limit of $f$ may or may not exist as in your examples.