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My question is about the following limit:

$$\large\lim_{x \to 0}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}$$

Where the notation $\lfloor x \rfloor$ is the floor function. I've graphed it and it is 1 everywhere except for $[0, 1)$. So, I think the answer should be 1, according to the epsilon-delta definition, since the function isn't defined for $0^+$, we can only check at $0^-$.

Is this correct? Also, is it to correct to simply cancel both the numerator and denominator here?

Thanks.

TheRandomGuy
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    Yes, the limit is $1$ by the $\varepsilon-\delta$ definition. You can cancel the numerator and denominator since they are equal and never $0$ in the domain of the function $x\mapsto\frac{\lfloor x\rfloor}{\lfloor x\rfloor}$. But remember that this function is not the constant function $x\mapsto1$ and don't confuse it with $x\mapsto1$ – Shubham Johri Dec 17 '18 at 09:47
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    Yes you are right, here the limits to $0$ and $0^-$ coincide. The fact that the limit to $0^+$ is undefined doesn't matter. –  Dec 17 '18 at 09:48
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    You may simplify num/denom, provided you restrict the domain of definition. –  Dec 17 '18 at 09:49
  • If you are addressing me, I did answer that. –  Dec 17 '18 at 09:52
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    The function is not defined for $0 <x<1$ so the question does not make sense. – Kavi Rama Murthy Dec 17 '18 at 09:54
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    @KaviRamaMurthy That depends on conventions. – egreg Dec 17 '18 at 09:54

2 Answers2

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It depends on how you define limits. If you require that a function is defined on a whole punctured neighborhood of $x_0$ (where you take the limit at), then this function doesn't have a limit at $0$.

If your definition is

Let $f$ be defined on a set $D$ and let $x_0$ be an accumulation point of $D$ ($x_0$ is not required to belong to $D$). We say that $$ \lim_{x\to x_0} f(x)=l $$ if, for every $\varepsilon>0$, there exists $\delta>0$ such that $$\text{for all $x\in D$, if $0<|x-x_0|<\delta$, then $|f(x)-l|<\varepsilon$}$$

then you can state correctly that $$ \lim_{x\to0}\frac{\lfloor x\rfloor}{\lfloor x\rfloor}=1 $$

Note 1. The condition that $x_0$ is an accumulation point of $D$ avoids problems with isolated points of the domain, so as to ensure uniqueness of the limit.

Note 2. With this definition, the limit from the right is the limit of the function $f$ restricted to the domain $D\cap(x_0,\infty)$. Similarly for the limit from the left.

egreg
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    $x_0$ is an accumulation point of $D-{x_0}$ is a redundancy. $x_0$ is an accumulation point of $D\iff x_0$ is an accumulation point of $D-{x_0}$, since accumulation points of sets aren't required to be in them – Shubham Johri Dec 17 '18 at 09:59
  • @ShubhamJohri You're right. – egreg Dec 17 '18 at 10:04
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Yes the function is defined only for $x<0$ and, according to the more general definition of limit, we have that

$$\lim_{x \to 0}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}=\lim_{x \to 0^-}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}=\lim_{x \to 0^-}1=1$$

For the different definition of limits refer also to the related

user
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