# 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?

• In order to be completely safe, I think it is better to say it is continuous from the right. Some here though think that continuous may be enough as far as we refer to points within the domain of definition. I don't like that but it may be a matter of agreement.
• $\lim_{x\to a} f(x)$ 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.
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