Floors?

[Wahamuka]

My teacher hasn’t discussed this yet so not sure where I’m headed

The greatest integer function is defined by g(x) = ⌊x⌋, where ⌊x⌋ is the largest integer that is less than or equal to x.
Show that lim x→2 g(x) does not exist.

Why does it not exist? Isn’t the answer 2 when you substitute it to find the limit?

I haven’t learned the concept of a derivative so don’t go there please haha

 

[pka]

The greatest integer function is defined by g(x) = ⌊x⌋, where ⌊x⌋ is the largest integer that is less than or equal to x.
Show that lim x→2 g(x) does not exist.

\(\displaystyle f(x)=\left\lfloor x \right\rfloor\) now \(\displaystyle f(1.9)=1,~f(2)=2~\&~f((2.1)=2\). Do you understand those evaluations?
Here is the crux of your question: \(\displaystyle \mathop {\lim }\limits_{\bf{x \to {2^ - }}} f(x) = 1\) BUT \(\displaystyle \mathop {\lim }\limits_{\bf{x \to {2^+}}} f(x) =2\)
The limit from the left is not the limit on the right.

 

[Dr.Peterson]

You don't find a limit by substituting. That should be one of the first things you learned about limits! Unfortunately, many of your first examples of proving limits may appear to contradict that, because you and up substituting -- in a continuous function, which is the one case in which you can do so.

This is not a continuous function. Just look at the graph. What happens as x approaches 2 from above? How about from below?

In general, to understand limits, you need to focus on the definition of a limit.

 

[Wahamuka]

\(\displaystyle f(x)=\left\lfloor x \right\rfloor\) now \(\displaystyle f(1.9)=1,~f(2)=2~\&~f((2.1)=2\). Do you understand those evaluations?
Here is the crux of your question: \(\displaystyle \mathop {\lim }\limits_{\bf{x \to {2^ - }}} f(x) = 1\) BUT \(\displaystyle \mathop {\lim }\limits_{\bf{x \to {2^+}}} f(x) =2\)
The limit from the left is not the limit on the right.

Ah yes! Thank you for making this clear, I didn’t think about testing it like this even though we were taught to do this (now that I recall).

Edit: wait... I’m confused how you found out that the lim x -> 2^- f(x) = 1. Why is it different from the limit coming from the right? Can’t seem to imagine it in my head. Can you give me an example.
Edit 2: Is it because the limit from the left is somewhere below 2 like 1.99999, so since we need to find the floor its 1? I’m pretty sure thats the reason.

You don't find a limit by substituting. That should be one of the first things you learned about limits! Unfortunately, many of your first examples of proving limits may appear to contradict that, because you and up substituting -- in a continuous function, which is the one case in which you can do so.

This is not a continuous function. Just look at the graph. What happens as x approaches 2 from above? How about from below?

In general, to understand limits, you need to focus on the definition of a limit.

Could you clarify what you mean by continuous function? How and why does this affect the concept of finding a limit. (My most recent lesson in school is “infinite limits and limits at infinity”)

 

[Wahamuka]

Oh, follow up question

Find all values of a for which lim x -> a g(x) exists.

For this my answer is all real except whole numbers. My taught process is that if its a whole number then it is sure that the limits from the left and right will never be the same, but for fractions for example they can.

 

[JeffM]

Ah yes! Thank you for making this clear, I didn’t think about testing it like this even though we were taught to do this (now that I recall).

Edit: wait... I’m confused how you found out that the lim x -> 2^- f(x) = 1. Why is it different from the limit coming from the right? Can’t seem to imagine it in my head. Can you give me an example.
Edit 2: Is it because the limit from the left is somewhere below 2 like 1.99999, so since we need to find the floor its 1? I’m pretty sure thats the reason.



Could you clarify what you mean by continuous function? How and why does this affect the concept of finding a limit. (My most recent lesson in school is “infinite limits and limits at infinity”)

A continuous function is continuous at every point in its domain (with some provisos at boundary points of closed intervals).

[MATH]\text {The real function } f(x) \text { is continuous at real } a \iff [/MATH]
[MATH]f(a) \in \mathbb R,\ \lim_{x \rightarrow a} f(x) \in \mathbb R, \text { and } f(a) = \lim_{x \rightarrow a} f(x). [/MATH]
But what do we mean by that limit?

[MATH]\lim_{x \rightarrow a^+} f(x) \in \mathbb R \text { and } \lim_{x \rightarrow a^-} f(x) = \lim_{x \rightarrow a^+} f(x) \iff[/MATH]
[MATH]\lim_{x \rightarrow a^+} f(x) = \lim_{x \rightarrow a} = \lim_{x \rightarrow a^-} f(x).[/MATH]
Pretty much the whole theory of differential calculus under standard analysis depends on finding the limit of

[MATH]\dfrac{f(x + h) - f(x)}{h}[/MATH]
as h approaches zero. Clearly, that Newton quotient does not exist at h = 0. So you cannot substitute h = 0. This is why they are pounding all these careful distinctions into your head.

I personally have always believed there is a better way to develop calculus, but am WAY too lazy to try to do it.

 

[JeffM]

Oh, follow up question

Find all values of a for which lim x -> a g(x) exists.

For this my answer is all real except whole numbers. My taught process is that if its a whole number then it is sure that the limits from the left and right will never be the same, but for fractions for example they can.

Perfectly good logic. You can always find a delta that keeps you within (integer, integer + 1).

 

[MarkFL]

For this my answer is all real except whole numbers.

Replace "whole numbers" with "integers." You could give this as:

[MATH]\{x\,|\,x\notin\mathbb{Z}\}[/MATH]
Whole numbers, as least as I was taught, are those integers which are not negative. But we want to exclude all integers.