Limits of the Greatest Integer Function

**mojo0716** · March 17th 2008, 05:55 AM

lim
x -> 3- (2-[[-x])

Why does this end up as (2 - - 3)(according to textbook) instead of (2 - - 2)?

Shouldn't I pick values of x less than 3, like 2.9 which would evaluate to 2 from the greatest integer function?

thanks

**wingless** · March 17th 2008, 07:23 AM

$\lim_{x\to 3^{-}} 2 - \lfloor -x \rfloor$

You can write 3^{-} as $3 + 0^{-}$ and treat $0^{-}$ as a negative number. For example, $-(0^{-}) = 0^{+}$ ..

$2 - \lfloor -(3^{-}) \rfloor$

$2 - \lfloor -(3 + 0^{-}) \rfloor$

$2 - \lfloor -3 - 0^{-}) \rfloor$

$2 - \lfloor -3 + 0^{+}) \rfloor$

$2 - \lfloor (-3)^{+}) \rfloor$

$2 - (-3) = 5$

Or we can solve it without using algebra.

$\lim_{x\to 3^{-}} 2 - \lfloor -x \rfloor$

The limit of $2 - \lfloor -x \rfloor$ as x approaches to a value a little less than 3.
Then $-x$ will be a little more than -3
So $\lfloor$ a little more than -3 $\rfloor = -3$
$2 - (-3) = 5$ .

**mojo0716** · March 17th 2008, 08:16 AM

Originally Posted by wingless

$\lim_{x\to 3^{-}} 2 - \lfloor -x \rfloor$

You can write 3^{-} as $3 + 0^{-}$ and treat $0^{-}$ as a negative number. For example, $-(0^{-}) = 0^{+}$ ..

$2 - \lfloor -(3^{-}) \rfloor$

$2 - \lfloor -(3 + 0^{-}) \rfloor$

$2 - \lfloor -3 - 0^{-}) \rfloor$

$2 - \lfloor -3 + 0^{+}) \rfloor$

$2 - \lfloor (-3)^{+}) \rfloor$

$2 - (-3) = 5$

Or we can solve it without using algebra.

$\lim_{x\to 3^{-}} 2 - \lfloor -x \rfloor$

The limit of $2 - \lfloor -x \rfloor$ as x approaches to a value a little less than 3.
Then $-x$ will be a little more than -3
So $\lfloor$ a little more than -3 $\rfloor = -3$
$2 - (-3) = 5$ .

Thanks for replying. I still don't understand it though.

According to the the textbook solution, the answer for

the limit from the right is 6

Actually, I have the solutions guide to this textbook, but it doesn't have every step

Here is what it

says

$\lim_{x\to 3} (2 - \lfloor -x \rfloor)$ does not exist - The two sides are not equal, so I understand this part.

because

$\lim_{x\to 3^{-}} (2 - \lfloor -x \rfloor)$ = 2 -(-3) = 5

and

$\lim_{x\to 3^{+}} (2 - \lfloor -x \rfloor)$ = 2 -(-4) = 6

Can you please explain the steps involved in reaching these answers?

thanks

**Plato** · March 17th 2008, 08:26 AM

Don't over think this problem.
Consider the following:
$\eta \in (4,5) \Rightarrow \quad \left\lfloor \eta \right\rfloor = 4\quad \& \quad - \eta \in ( - 5, - 4) \Rightarrow \quad \left\lfloor { - \eta } \right\rfloor = - 5.<br />$

**mojo0716** · March 17th 2008, 08:42 AM

Originally Posted by Plato

Don't over think this problem.
Consider the following:
$\eta \in (4,5) \Rightarrow \quad \left\lfloor \eta \right\rfloor = 4\quad \& \quad - \eta \in ( - 5, - 4) \Rightarrow \quad \left\lfloor { - \eta } \right\rfloor = - 5.<br />$

Ok, I have no clue what that means. I am not familiar with that notation. I have not seen that yet. I am taking Calculus for the first time and in a quarter

We are starting with limits, which I have heard is precalculus. I have had College Algebra and Trig so far.

thanks

**Plato** · March 17th 2008, 08:59 AM

Originally Posted by mojo0716

I am not familiar with that notation. I have not seen that yet.

1) Are you saying that you do not understand the greatest integer function?
2) Or are saying that you do not understand the symbol that both wingless and I used to represent the greatest integer function?

It has come to be common to use $\left\lfloor \eta \right\rfloor$ whereas older texts use $\left[ {\left[ \eta \right]} \right]$ .

If it is case (1) then you need PreCalculus.
If it is just a matter of symbols I think we can straighten that out.

**mojo0716** · March 17th 2008, 09:23 AM

Originally Posted by Plato

1) Are you saying that you do not understand the greatest integer function?
2) Or are saying that you do not understand the symbol that both wingless and I used to represent the greatest integer function?

It has come to be common to use $\left\lfloor \eta \right\rfloor$ whereas older texts use $\left[ {\left[ \eta \right]} \right]$ .

If it is case (1) then you need PreCalculus.
If it is just a matter of symbols I think we can straighten that out.

I think I understand the Greatest Integer function. There is no pre-calculus class at my college. College Algebra is supposedly most of precalculus(I believe no limits though). Transformations of functions, functions, logarithms, asymptotes are some of the topics in that class I have completed.

The Greatest Integer function is also called the floor function? I am a computer science major and the floor function is used in a lot of programming languages, so I am somewhat familiar with it. It basically rounds down to the closest integer? e.g. 1.5 = 1, 2.9 = 2, -2.5 = -3?

I have never seen some of the symbols you have used. I do know the symbol for the greatest integer function = [x]. I don't know the funny looking n and e characters(eta?) symbolize though.

The way I see this problem is that as x approaches 3 from the left, you can pick something like 2.99, and that evaluates to 2 using the floor function?

When x approaches 3 from the right, you can pick something like 3.0001, which evaluates to 3 from the floor function?

I don't know how the book got 3 from the left and 4 from the right. I would think it would be 2 from the left and 3 from the right.

I don't know what I am doing wrong. I appreciate the help and I hope you understand what I am saying.

Thanks

thanks

**Plato** · March 17th 2008, 09:44 AM

O.K. then I will use the floor function notation (which by the way comes from computer science into mathematics).
If $x \to 3^ +$ it is a real number between 3 and 4, $x \in (3,4) \Rightarrow \quad \left\lfloor x \right\rfloor = 3$ .
If x is a real number between 3 and 4 then –x is is a real number between -4 and –3, $- x \in ( - 4, - 3) \Rightarrow \quad \left\lfloor { - x} \right\rfloor = - 4$ .
Thus as $<br /> x \to 3^ + ,\,\left[ {\left( {2 - \left\lfloor -x \right\rfloor } \right)} \right] \approx \left[ { \left( {2 - \left( { - 4} \right)} \right)} \right]$
So the right hand limit is 6.

**mojo0716** · March 17th 2008, 09:51 AM

Originally Posted by Plato

1) Are you saying that you do not understand the greatest integer function?
2) Or are saying that you do not understand the symbol that both wingless and I used to represent the greatest integer function?

It has come to be common to use $\left\lfloor \eta \right\rfloor$ whereas older texts use $\left[ {\left[ \eta \right]} \right]$ .

If it is case (1) then you need PreCalculus.
If it is just a matter of symbols I think we can straighten that out.

Originally Posted by Plato

O.K. then I will use the floor function notation (which by the way comes from computer science into mathematics).
If $x \to 3^ +$ it is a real number between 3 and 4, $x \in (3,4) \Rightarrow \quad \left\lfloor x \right\rfloor = 3$ .
If x is a real number between 3 and 4 then –x is is a real number between -4 and –3, $- x \in ( - 4, - 3) \Rightarrow \quad \left\lfloor { - x} \right\rfloor = - 4$ .
Thus as $<br /> x \to 3^ + ,\,\left[ {\left( {2 - \left\lfloor x \right\rfloor } \right)} \right] \approx \left[ { \left( {2 - \left( { - 4} \right)} \right)} \right]$
So the right hand limit is 6.

Ok that makes a lot more sense to me. Thanks to both of you for the help!

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