[Math] A limit exists iff and only the left limit and the right limit exist and are equal to each other

Question

It is well known that

$$\lim_{x \to a} f(x) = L \iff \lim_{x \to a+}f(x) = L = \lim_{x \to
a-}f(x)$$

Consider the function $\sqrt{.}: \mathbb{R}^+ \to \mathbb{R}$

Now, consider $\lim_{x \to 0} \sqrt{x}$

We can prove this limit is equal to $0$. Indeed, let $\epsilon > 0$. Choose $\delta = \epsilon^2$. Then, for $x \in \mathbb{R}^+$ satysfying $0 < |x| < \delta$ or equivalently $0 < x < \delta$, we have $\sqrt{x} < \sqrt{\delta} = \epsilon$, which establishes the result.

However, my confusion lies in the following: the limit from the left does not seem to exist, making the above theorem untrue. Where lies my mistake?

Answer 1

Actually, what you wrote is not well known, and is in fact not even true.

What is well known is the fact that

Let $f$ be a function $\color{red}{\text{defined on some open neighborhood of $a$}}$. Then, $$\lim_{x \to a} f(x) = L \iff \lim_{x \to a+}f(x) = L = \lim_{x \to a-}f(x)$$

Always read the fine print.