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The $\lim_{x\to 0} \frac{\sqrt{x}}{x}$ can be easily evaluated by simplification: $\lim_{x\to 0} \frac{\sqrt{x}}{x} = \lim_{x\to 0} \frac{1}{\sqrt{x}}$. At this point, the right-hand limit can be taken: $\lim_{x\to 0+} \frac{1}{\sqrt{x}}=+\infty$, but left-hand limit $\lim_{x\to 0-} \frac{1}{\sqrt{x}}$ can not be evaluated as the function is not defined for $x < 0$. So my question now is: $\lim_{x\to 0} \frac{\sqrt{x}}{x} = +\infty$ or should I say it does not exist? My doubt comes from the fact that $\lim_{x\to0} \sqrt{x}=0$, as explained in this (answer).

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In order that $\sqrt x$ is defined you have to take the domain as $[0,\infty)$ and so you can only talk about the right-hand limit. The required limit is $+\infty$.

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  • $\begingroup$ So should I say that both $\lim_{x \to 0} \sqrt{x}$ and $\lim_{x \to 0} \frac{\sqrt{x}}{x}$ only exit for the restricted domain $[0, \infty]$? I was convinced (link answer in the question) that $\lim_{x \to 0} \sqrt{x}=0$ because we can not evaluate the function for $x<0$, then we can not apply the left and right-hand theorem. $\endgroup$ – Marcos Alex Jul 6 at 23:22
  • $\begingroup$ The first sentence is correct it infinite limits are allowed. If you want to consider only finite limits then none of the limits exist. $\endgroup$ – Kabo Murphy Jul 6 at 23:25
  • $\begingroup$ What do you mean by taking the domain as $\mathbb{R}$ (or $[0, \infty)$ for that matter)? The function $f(x)=1/\sqrt{x}$ is (if we're talking real analysis) only defined for $x > 0$. $\endgroup$ – Hans Lundmark Jul 7 at 8:06
  • $\begingroup$ @HansLundmark I have edited the answer. $\endgroup$ – Kabo Murphy Jul 7 at 12:23
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Note that you can think of: $$\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$$ and so: $$\lim_{x\to0^+}\frac{\sqrt{x}}{x}=\lim_{x\to0^+}\frac{1}{\sqrt{x}}\to\infty$$

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The limit, in order to exist, must be two-sided. As $\lim_{x \to 0^{-}}\frac{\sqrt{x}}{x}$ DNE, it follows that $\lim_{x \to 0} \frac{\sqrt{x}}{x}$ DNE. (However, the RHL is equal to $\infty$ as Henry Lee points out.)

Similarly, $\lim_{x \to 0} \sqrt{x}$ DNE because the LHL does not exit; however, the RHL is equal to $0$.

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    $\begingroup$ “The limit, in order to exist, must be two-sided”: what world authority has so decreed? $\endgroup$ – egreg Jul 7 at 14:18
  • $\begingroup$ LHL and RHL must exist and be equal; otherwise the limit as we define it does not exist; that is, a limit is ``two-sided.'' $\endgroup$ – mlchristians Jul 7 at 15:17
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    $\begingroup$ Who is “we”? I'm surely not among them. If a function $f$ is defined on a set $D$ and $c$ is a limit point of $D$, then I (and am not the only one) say that $\lim_{x\to c}f(x)=L$ if (and only if) for every $\varepsilon>0$ there exists $\delta>0$ with the following property: for every $x\in D$, if $0<|x-c|<\delta$, then $|f(x)-L|<\varepsilon$. An obvious consequence of this definition is that $\lim_{x\to0}\sqrt{x}=0$. $\endgroup$ – egreg Jul 7 at 15:22
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    $\begingroup$ Did you read carefully, including the condition $x\in D$? Again, your we does not include me and many other people. $\endgroup$ – egreg Jul 7 at 15:51
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    $\begingroup$ OK, then the function $\sqrt{x}$ is not continuous at zero, is it? $\endgroup$ – egreg Jul 7 at 16:19

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