If I'm not wrong, $\lim_ {x \to 0} \sqrt{x} = 0$ , even though $\lim_ {x \to 0^{-}} \sqrt{x}$ does not exist.

Then in this following graph of $f(x)$ :

enter image description here

Does $\lim_{x \to 1} f(x)$ exist? Would it still exist if $f(1)$ were not defined?

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Since $\sqrt x$ is undefined when $x<0$, it doesn't make sense to talk about $\lim_{x\to0^-}\sqrt x$.

And, yes, $\lim_{x\to1}f(x)=1$, even if $f(1)$ is undefined.

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  • 1
    $\begingroup$ I don't think $\;\lim\limits_{x\to1}f(x)\;$ is even defined as the function isn't defined on some left neighborhood of $\;x=1\;$ ... This is completely similar to $\;\lim\limits_{x\to0}\sqrt x\;$ . $\endgroup$ – DonAntonio Apr 18 '18 at 9:42
  • $\begingroup$ @DonAntonio I don't understand you comment. It is defined in some interval $(1,1+\varepsilon)$. $\endgroup$ – José Carlos Santos Apr 18 '18 at 9:43
  • $\begingroup$ The same you could say about $\sqrt x$ : is is defined on the whole ray $\;[0,\infty)\;$ ...why is this case different? The limit is assumed to be taken on both sides ... What we can talk about is $\;\lim\limits_{x\to1^+} f(x)\;$ . $\endgroup$ – DonAntonio Apr 18 '18 at 10:01
  • $\begingroup$ In my opinion, in order to speak about $\lim_{x\to a}f(x)$, all we need is that $a$ is an accumulation point of the domain $D_f$ of $f$. But in order to talk about $\lim_{x\to a^+}f(x)$, $a$ must be an accumulation point of $D_f\cap(a,+\infty)$ and in order to talk about $\lim_{x\to a^-}f(x)$, $a$ must be an accumulation point of $D_f\cap(-\infty,a)$. $\endgroup$ – José Carlos Santos Apr 18 '18 at 10:10
  • $\begingroup$ Accumulation point is a notion from basic topology which is, as far as I know, introduced later in analysis. As far as I know, it is usually stressed that $\;\lim\limits_{x\to a} f(x)=A\;$ if no matter how $\;x\;$ approaces $\;a\;$ ...etc. The absolute value in the definition also supports this view, otherwise we could begin with one-sided limits and then generalize the definition... $\endgroup$ – DonAntonio Apr 18 '18 at 15:10

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