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Math Help - Huge headache on evaluating floor function limits

  1. #1
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    Huge headache on evaluating floor function limits

    I have a few questions regarding floor functions and how to evaluate their limits:

    So far, when asked to evaluate the limit of an expression's floor function, I would simply get the floor function of x and substitute this value for the expression.

    For example given the problem:

    lim [_x+1_] as x-> 3 from the left,

    I know the function splits, giving me
    2 <= x < 3 (approaching 3 from the left)
    3 <= x < 4 (approaching 3 from the right)

    therefore, I take 2 and plug it in to the function, giving me a result of 3. This is consistent with wolfram alpha's answer
    http://tinyurl.com/ycq75av

    However, I get the wrong answer when I do the same exact thing in:
    lim [_2x-1_] as x-> -2 from the left
    - Wolfram|Alpha

    What I do is like what I have shown above, that is:
    -3 <= x < -2
    -2 <= x < -1

    So I take -3, plug it in to (2x-1) and get (2)(-3) - 1= (-6) - 1 = -7 when the correct answer is given as -6.

    The method I'm using works when I have to get the limit as x approaches -2 from the right though. So given these kinds of problems, what's the proper way of evaluating floor functions?

    On the other hand, when I'm given a function
    lim [_1-x_] as x-> -1 from the left, I get the intervals

    -2 <= x < -1
    -1 <= x < 0

    Plugging in the value of -2 yields 3 when the correct answer is 2 while solving for the same function as x approaches -1 from the right should yield 1, but plugging in -1 will result in 2.

    I was able to evaluate the limit as x-> -1 from the left by doing it this way:
    -2 <= 1-x < -1
    -3 <= -x < -2
    3 > x >= 2

    and getting the consistent result of 2

    But if I do the same procedure to any of the previous examples, I get the wrong result. For example:

    lim [_x+1_] as x-> 3 from the left

    2 <= x+1 < 3
    1 <= x < 2

    Why does this happen?

    Last question:
    lim [_2x-3_] as x->-2 from the left

    According to wolfram alpha the answer is -8
    - Wolfram|Alpha

    Now I'm really confused as to how this answer came up. I can get -9 via
    -3 <= x < -2, getting -3 and substituting 2(-3) - 3 = -9. Where did -8 come from?

    Any help regarding any of my questions would be greatly appreciated, especially with regard to the proper steps in evaluating floor functions in general.

    On another note, is it possible that wolfram alpha is wrong?
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  2. #2
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    [x]=the greatest integer less than x .
    say x=3.111 then [x]=3
    say x=-3.111 then [x]=-4
    say x=3 then [x]=3
    say x=-3 then [x]=-3

    i couldnot figure out how the function splits. but i have another method to solve it. we know that [x]=x-{x} where {x}=constant as 0<{x}<1. so we break the functions like [2x-1]=2x-1-constant.putting limits we may get the required answer. also floor functions are generally not continuous. and unless a function is continuous u cannot find its limit.please comment!!
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  3. #3
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    Quote Originally Posted by Pulock2009 View Post
    and unless a function is continuous u cannot find its limit.please comment!!
    we talk about limit from the left, so the function need not to be continuous.

    \lim\limits_{x\longrightarrow3^{-}}\left\lfloor x+1\right\rfloor
    we want to know the limit when x approach 3 from the left, so let x=3-\varepsilon, \varepsilon>0,
    and we get
    \left\lfloor x+1\right\rfloor= \left\lfloor (3-\varepsilon)+1\right\rfloor= \left\lfloor 4-\varepsilon\right\rfloor
    from the definition of floor function, we get
    \left\lfloor 4-\varepsilon\right\rfloor= 3.

    \lim\limits_{x\longrightarrow-2^{-}}\left\lfloor 2x-3\right\rfloor
    using the same method, let x=-2-\varepsilon, \varepsilon>0,

    hence,

    \left\lfloor 2x-3\right\rfloor = \left\lfloor 2(-2-\varepsilon)-3\right\rfloor = \left\lfloor -7-2\varepsilon\right\rfloor = -8
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  4. #4
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    Quote Originally Posted by dedust View Post
    we talk about limit from the left, so the function need not to be continuous.

    we want to know the limit when x approach 3 from the left, so let x=3-\varepsilon, \varepsilon>0,
    and we get
    \left\lfloor x+1\right\rfloor= \left\lfloor (3-\varepsilon)+1\right\rfloor= \left\lfloor 4-\varepsilon\right\rfloor
    from the definition of floor function, we get
    \left\lfloor 4-\varepsilon\right\rfloor= 3.


    using the same method, let x=-2-\varepsilon, \varepsilon>0,

    hence,

    \left\lfloor 2x-3\right\rfloor = \left\lfloor 2(-2-\varepsilon)-3\right\rfloor = \left\lfloor -7-2\varepsilon\right\rfloor = -8
    I'm sorry this went way over my head. What exactly is happening here? Is epsilon 0<e<1 or merely greater than 0? Can it be greater than or equal to 0? What about 1? From pulock's post, it seems that you're using the same way of evaluating as he is.

    From what I understand, you define x as being [(value x is approaching) - (constant between 0 and 1)] and then you plug in any number between 0 and 1 for epsilon, and get the resulting floor function. Am I correct?

    Using your method for lim [[5-2x]] as x->1 from the left doesn't give me the correct result though.

    let x = 1-e
    [[5-(2)(1-e)]]
    [[5-2+2e]]
    [[3+2e]]
    If i let e be , say .5 then

    [[3+2(.5)]]
    [[3+1]]=4 when the answer should be 3.

    How do you solve the above problem from the left and right?

    Is there a method of evaluating that doesn't make use of subtracting a constant?

    Also, what's the procedure if I'm getting limits from the right? Is it: value x is approaching + 0<e<1 ?
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  5. #5
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    \varepsilon is approaching 0, not a fixed value.
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  6. #6
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    Quote Originally Posted by dedust View Post
    \varepsilon is approaching 0, not a fixed value.
    Hmm alright, but how does one go about eliminating e to get a numerical value? What can you say about my previous observations to the method in my previous post? Can you clear it up a bit/correct it?

    Thanks btw, great help so far, just having a hard time digesting it
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