Difference between these two limits I want to know the difference between the following limits:
$$\lim_{x\to 1^{+}} \frac{1}{\lfloor{x}\rfloor-1},$$
$$\lim_{x\to 1^{+}} \frac{1}{x-1}$$
Is the answer to both infinite?
What is the difference between undefined and infinite?
 A: Short answer: Some calculus textbooks say that an infinite limit is undefined, so there is no difference between the two, while others say infinite limits are defined, so there is a difference.

Longer answer:
In the first limit, if $x$ is close to $1$ but to the right of it  then $\lfloor x\rfloor=1$, so the expression inside the limit is $\frac 10$ and is undefined. That means the limit is completely undefined.
In the second limit, the expression inside the limit is defined for $x$ close to $1$ but to the right of it. However, as $x\to1^+$, the expression increases without bound.
There is therefore no real number that is the second limit. Some calculus textbooks say that the limit is undefined.
However, some calculus textbooks say that the limit is defined and equals $+\infty$. In other words, they agree that the limit is not a real number, but they add two more "numbers" so some more limits exist. The real numbers plus the "numbers" $+\infty$ and $-\infty$ are called the "affine extended real number system." Looking at limits this way has the advantage that you can talk about differences between the two kinds of limit you show. There is also the disadvantage that some theorems that are true with finite limits are not true if you allow infinite limits, so the theorems must be more carefully stated.
I have seen and used both kinds of calculus textbooks, and both are common. I cannot recommend one kind over the other, in general. The two philosophies do sometimes make some questions ambiguous, as in your question. The calculus textbook I now use to teach sticks infinite limits into one skippable section, so outside that section it treats infinite limits as undefined. In another section it does explain multiple causes for limits to be undefined and would distinguish your two kinds of causes (as well as others such as oscillations).
A: $$\lim_{x \to 1^+} {1 \over {\lfloor x \rfloor-1}}={1 \over \lfloor 1+\epsilon \rfloor -1}$$
where epsilon is for this intent and purpose a positive small number, possibly arbitrary small, maybe infinitesimal. The floor of any number between 1 and 2, however, is 1 so we end up with.
$${1 \over {1-1}}={1 \over 0}=sign(0) \cdot \omega$$
where $\omega= {1 \over {\epsilon}}$
Clearly the limit doesn't exist, unless you like complex infinity, but I assigned it a value anyway. Usually you'll hear $1 \over 0$ be considered undefined, but that's not really the most we can say about the value.
As for the second limit...
$$\lim_{x \to 1^+} {1 \over {x-1}}={1 \over {1+\epsilon-1}}={1 \over {\epsilon}}=\omega$$
better known as $\infty$
So what's the difference? In the first limit, the denominator can only take on discrete values, this prevents the limit from existing. In the second limit, the value of zero is approached in a smoother way allowing for the limit to exist. The difference in value being, the definiteness of the infinity. In the first, the value is every infinity that can exist. In the second, just positive infinity.
Perhaps I should note that having a one sided limit not exist is quite rare as far as what you'll see in practice. Usually if a regular limit doesn't exist, you'll just compare one sided limits. However, in this case, the secondary one sided limit doesn't even exist!
A: In the neighborhood of $1^+$,
$$\lfloor x\rfloor-1=0$$while$$x-1>0.$$
For this reason, the first limit is undefined ($1/0$ is not defined as we can't decide between $\pm\infty$) and the second is infinite ($1/0^+$ is $+\infty$). Some authors just say undefined in both cases.
A: If we go from definition. we can say that:
$$\lim_{x\to1^+}\frac{1}{x - 1} = \infty$$
because for all $\epsilon > 0$, there exists $\delta > 0$ such that $f(x) > \epsilon$ whenever $|x-1| < \delta$ (specifically pick $\delta = \frac1{\epsilon}$).
However, $\lim_{x\to1^+}\frac{1}{\lfloor{x}\rfloor - 1}$ does not exist. If we pick any $\epsilon > 1$, there can no such $\delta$ since $f(x) \ngtr 1$ for all $x \geq 2$ and undefined from $x \in [1,2)$.
A: On the interval $(1,2),$ the first function contains no values, and therefore a right-handed limit can not be determined.  I'm assuming then, the answer is not infinite, but rather undefined.
A: Good question.  We often think of the limit of a function as an object, but it's really a statement about the function and a point.  It tells you the behavior of the function near that point.  
We say that $\lim_{x\to 1^+} \frac{1}{x-1} = \infty$ because values of $\frac{1}{x-1}$ can be made arbitrarily large by taking $x$ sufficiently close to $1$ but greater than $1$.
However, there are no values of $f(x) = \frac{1}{\lfloor x\rfloor -1}$ when $x$ is close to $1$ but greater than $1$, for the reasons explained in other answers.  So it is meaningless to say anything about the function near $1$.  
Some will say $\lim_{x\to 1^+}\frac{1}{\lfloor x\rfloor -1}$ “is undefined,” but I think that can enforce the misconception that “undefined” is some kind of value a limit can take.  I think it's better to say that $\lim_{x\to 1^+}\frac{1}{\lfloor x\rfloor -1}$ does not exist.
