# Why is $f(x) = \sqrt{x}$ is continuous at $x=0$? Does $\lim_{x \to 0^{-}} \sqrt{x}$ exist?

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You make a single mistake in your question: you forget to specify the domain of definition of $\sqrt \cdot$. You know that this is $[0, \infty)$, so "limit in $0$" here means, necessarily, "limit from the right" - simply because there is nothing to the left of $0$ in $[0, \infty)$.

If you prefer to be more pedantic, you could say that $[0, \infty)$ carries the subspace topology induced by the topology of $\Bbb R$: a subset $U \subseteq [0, \infty)$ is open (in the topology of $[0, \infty)$) if and only if there exist some open subset $V \subseteq \Bbb R$ (in the topology of $\Bbb R$) such that $U = V \cap [0, \infty)$. It is enough to notice now that all these open subsets $U$ are "truncated to the left of $0$".

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### BCLC

Updated on August 16, 2022

If I remember right, $f(x)$ is continuous at $x=a$ if

1. $\lim_{x \to a} f(x)$ exists

2. $f(a)$ exists

3. $f(a) = \lim_{x \to a} f(x)$

So $\lim_{x \to 0^{-}} \sqrt{x}$ exists? Thus $\lim_{x \to 0^{-}} \sin(\sqrt{x})$ exists?

$\lim_{x\to a} f(x)$ does not care about anything outside the domain of $f$ (except possibly $a$, which must be an adherent point, or even an accumulation point of the domain, depends on the used definition of the limit). If your $\sqrt{x}$ is only defined for $x \geqslant 0$, then $\lim_{x\to 0} \sqrt{x} = \lim_{x \to 0^+} \sqrt{x}$. If the domain also contains negative real numbers … it should take complex values, but still be continuous at $0$.
Note that the definition of $\lim_{x\to a}f(x)$ makes no reference to $\lim_{x\to a^+}f(x)$ or $\lim_{x\to a^-}f(x)$. Those three concepts, while clearly similar, exist (almost) independently of one another.