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Does the limit $\lim_{x \to 1}\left(\frac{x}{\lfloor x\rfloor}\right)$ exist? Where $\lfloor x\rfloor$ is the Greatest Integer Function or the Floor function.

My teacher defined that a limit exists when the Left-hand Limit = Right-Hand Limit and If any one is undefined but the other is finite then also the limit exists. Like $\lim_{x \to 0}\sqrt{x}$ does exist although the LHL does not lie in the domain.

In this question the domain of the function $\frac{x}{\lfloor x\rfloor}$ is $\mathbb R\setminus[0,1)$ where $\mathbb R$ represents Real Numbers. So, The Left-Hand Limit is not defined and Right Hand Limit is $1$, which is finite so according to my teacher Limit does exist. But when I tried this question in WolframAlpha, it is saying that the LHL is $-\infty$ and RHL is $1$, so limit does not exist.

So, a more general question is: do we have to consider the domain of a function when we calculate a limit?

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Jan 18 at 8:16
  • $\begingroup$ An example of a limit that is not defined in the reals is $\lim_{x\to-1}\sqrt{x}$ is undefined in the reals, as for any $\epsilon$ that you choose that is less than $1$, $\sqrt{-1+\epsilon}$ and $\sqrt{-1-\epsilon}$ are undefined. But on the other hand, the right-hand limit $\lim_{x\to 0^+}\sqrt{x}$ is defined, as $\sqrt{0+\epsilon}$ is still defined for arbitrarily small positive $\epsilon$. $\endgroup$
    – C7X
    Jan 18 at 8:21
  • $\begingroup$ @C7X So $\lim_{x \to 0^+}(\sqrt{x})$ does exist but $\lim_{x \to 0}(\sqrt{x})$ does not as it is not in domain? $\endgroup$
    – Temp
    Jan 18 at 8:23
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    $\begingroup$ It could be rephrased as the continuity of sgn(x) at 1 $\endgroup$
    – RandomGuy
    Jan 18 at 8:25
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    $\begingroup$ @C7X My question has been answered that we have to consider the domain. Thanks for the effort $\endgroup$
    – Temp
    Jan 18 at 8:28

2 Answers 2

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Your teacher is completely right, according to the more general definition of limit, we consider $x$ in the domain and therefore

$$\lim_{x \to 1}\left(\frac{x}{[x]}\right)= \lim_{x \to 1^+}\left(\frac{x}{[x]}\right)=1$$

Refer also to the related

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  • $\begingroup$ Ok. So This means that we have to consider the domain for calculating the limit. Thanks $\endgroup$
    – Temp
    Jan 18 at 8:25
  • $\begingroup$ Yes, we are taking $x\in D\setminus\{1\}$ $\endgroup$
    – user
    Jan 18 at 8:27
  • $\begingroup$ It has to be VERY clear from the context, otherwise it is abuse of notation. I think it would be preferable to always write $\lim\limits_{\underset{\Large{x\in\mathcal D}}{x\to 1}}f(x)$ $\endgroup$
    – zwim
    Jan 18 at 12:09
  • $\begingroup$ @zwim I'm also thinking for example to $\lim_{x \to 0} x \log x$, for which we are implicitely assuming $x>0$ but it is not usually indicated as a right side limit. Also in the most autorative textbooks the definition of limit implies that $x$ is in the domain. We have discussed in detail this issue other times. Anyway in case of doubt, the notation you are proposing is a good notation. $\endgroup$
    – user
    Jan 18 at 12:17
  • $\begingroup$ @zwim This is a very complete reference for this kind of discussion math.stackexchange.com/q/2889055. I've incorporated it in the aswer. $\endgroup$
    – user
    Jan 18 at 12:21
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First of all to solve this question of limit first of all you have to be familiar with the definition of limit.

$$\lim_{x\to a} f(x)$$ will exist only when both the left hand and right hand limits will exist. That is $$\lim_{x\to a} f(x)$$ will exist only when $$\lim_{x\to a^{+}}f(x)$$ and $$\lim_{x\to a^{-}}f(x)$$ both will exist.

Now according to your question $$\lim_{x\to 1}(\frac{x}{[x]})$$ will exist only when $$\lim_{x\to 1^{+}}(\frac{x}{[x]})$$ and $$\lim_{x\to 1^{-}}(\frac{x}{[x]})$$ both will exist. Now the interesting fact is that $$\lim_{x\to 1^{+}} [x] = 1$$ and $$\lim_{x\to 1^{-}} [x] = 0$$. So, as far as I can guess $$\lim_{x\to 1^{-}}(\frac{x}{[x]})$$ will be undefined. Therefore since $$\lim_{x\to 1^{-}}(\frac{x}{[x]})$$ is undefined so, $$\lim_{x\to 1}(\frac{x}{[x]})$$ will not exist.

Though I am not sure whether my logic is correct or not. But I would suggest to verify once.

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  • $\begingroup$ this column presentation is not very appealing, can you consider something less chopped. $\endgroup$
    – zwim
    Jan 18 at 11:53
  • $\begingroup$ You can edit my answer and write it in a good manner@zwim $\endgroup$ Jan 18 at 13:42

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