I am trying to find the necessary $\delta$ that satisfies the $\epsilon – \delta$ definition of continuity of the following function at $x = 0$:
$$f(x) = \begin{cases} x & \text{ for } x \ge 0,\\
x^2 & \text{ for } x < 0.\end{cases}$$
My work:
right limit:
$\lim\limits_{x \to 0+}x$ $\implies \forall \epsilon$ $\exists \delta $ ($|x|<\delta$ $\implies$ $|f(x)|< \epsilon)$
clearly we can let $\delta = \epsilon$ as $f(x) = x$
For the left limit:
$\lim\limits_{x \to 0-}x^2$ $\implies \forall \epsilon$ $\exists \delta $ ($|x|<\delta$ $\implies$ $|f(x)|< \epsilon)$
here we let $\delta = \sqrt\epsilon$ as $f(x) = x^2$
My problem is that which $\delta$ do I choose?
Best Answer
Once you've found one value $\delta_1 > 0$ and used it to prove a statement of the form
then you can replace $\delta_1$ by any value $\delta$ such that $0 < \delta \le \delta_1$ and it will still be true that
Using that, suppose you have two values $\delta_1 > 0$ and $\delta_2 > 0$, and you have proved two statements like this:
$|x| < \delta_1 \implies P(x)$
$|x| < \delta_2 \implies Q(x)$
It follows that you can replace each of $\delta_1$ and $\delta_2$ by the same value $\delta = \min\{\delta_1,\delta_2\}$, and then you'll know its true that
In your problem, you have $\delta_1 = \epsilon$ and $\delta_2 = \sqrt{\epsilon}$, and so you use $\delta = \min\{\epsilon,\sqrt{\epsilon}\}$.