# Does $\lim_{x\to 0} \frac{\sqrt{x}}{x}$ exist? A comparison with the $\lim_{x\to 0} \sqrt{x}$

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).

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 $$+\infty$$.

• 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 6 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. – Kabo Murphy Jul 6 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 7 at 8:06
• @HansLundmark I have edited the answer. – Kabo Murphy Jul 7 at 12:23

Note that you can think of: $$\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$$ and so: $$\lim_{x\to0^+}\frac{\sqrt{x}}{x}=\lim_{x\to0^+}\frac{1}{\sqrt{x}}\to\infty$$

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 $$0$$.

• “The limit, in order to exist, must be two-sided”: what world authority has so decreed? – egreg Jul 7 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.'' – mlchristians Jul 7 at 15:17
• 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 7 at 15:22
• Did you read carefully, including the condition $x\in D$? Again, your we does not include me and many other people. – egreg Jul 7 at 15:51
• OK, then the function $\sqrt{x}$ is not continuous at zero, is it? – egreg Jul 7 at 16:19