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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).

N. F. Taussig
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Marcos Alex
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3 Answers3

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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 $+\infty$.

Kavi Rama Murthy
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  • 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. – Marcos Alex Jul 06 '19 at 23:22
  • The first sentence is correct it infinite limits are allowed. If you want to consider only finite limits then none of the limits exist. – Kavi Rama Murthy Jul 06 '19 at 23:25
  • 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 $x > 0$. – Hans Lundmark Jul 07 '19 at 08:06
  • @HansLundmark I have edited the answer. – Kavi Rama Murthy Jul 07 '19 at 12:23
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Note that you can think of: $$\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$$ and so: $$\lim_{x\to0^+}\frac{\sqrt{x}}{x}=\lim_{x\to0^+}\frac{1}{\sqrt{x}}\to\infty$$

Henry Lee
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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 $0$.

DDS
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    “The limit, in order to exist, must be two-sided”: what world authority has so decreed? – egreg Jul 07 '19 at 14:18
  • 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.'' – DDS Jul 07 '19 at 15:17
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    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 $\lim_{x\to0}\sqrt{x}=0$. – egreg Jul 07 '19 at 15:22
  • With all due respect, I don't know what Calculus book you are teaching from, but the idea of the $\delta-\epsilon$ definition you allude to means that we can get arbitrarily close to $0$ (in the example you cite) from both sides, As this is impossible in the case of $f(x) = \sqrt{x},$ the RHL exists and is equal to $0$. The LHL does not exist; and therefore, the limit as $we$ know it by the $\delta-\epsilon$ definition does not exist. It is perhaps because of arguments like the one that you are trying to make that this definition of limit took over two thousand years to get to us. – DDS Jul 07 '19 at 15:48
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    Did you read carefully, including the condition $x\in D$? Again, your we does not include me and many other people. – egreg Jul 07 '19 at 15:51
  • I never implied it did; but it does include authors from widely accepted Calculus textbooks, including the ones I have often had access to. – DDS Jul 07 '19 at 16:03
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    OK, then the function $\sqrt{x}$ is not continuous at zero, is it? – egreg Jul 07 '19 at 16:19
  • I'm glad you ask that. The function $\sqrt{x}$ is continuous over $(0, \infty)$ and continuous from the right at $0$. I don't mean to be presumptuous, but based on what you have previously argued, you would have to say it is continuous at $0$ and thus continuous over $[0, \infty]$. But without the distinction that I have tried to bear witness to, why do such notions as right-hand and left-hand limits, as well as the notions of continuity from the right and continuity from the left appear in textbooks? Certainly, they are there for reasons that pertain to our discussion. – DDS Jul 07 '19 at 18:49
  • @mlchristians: Surely every mathematician agrees that the (real) square root function is continuous, which includes it being continuous at the point $x=0$. If the function is defined for $x>0$ and undefined for $x<0$, then (as egreg is correctly trying to point out to you) the limit as $x \to 0$ is the same thing as the right-hand limit as $x \to 0^+$. See the similar recent discussion here, for example: https://math.stackexchange.com/questions/3279554/what-is-the-point-0-0-of-sqrtx-called – Hans Lundmark Jul 07 '19 at 19:33