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Why does cancellation in limits work like this?

  1. Jul 12, 2012 #1

    There isn't a problem statement, I'm just confused about something.

    2. Relevant equations
    3. The attempt at a solution

    Ok, for a while I didn't understand how cancelling in the following limit works:

    lim ----------
    x--> 5 (x-5)

    where you cancel the (x-5)s, because it looked like dividing by zero. Then I (think) I understood correctly, which is that, given f(x), the limit x--> 5 is totally different from f(5). In the latter case you are dividing by zero and in the limit, you are not dividing by zero, the x only approaches 5. Therefore you can cancel the (x-5)'s.

    I really hope I've understood at that correctly, if not please explain it to me. Now, here's the bit I REALLY don't understand (:D) :

    If I have a limit like:

    lim x--> 1, 2x(x-1)

    an online limit calculator tells me the answer is 4. I can also understand how it got to 4,
    (take the 2 out, multiply both the numerator and the denominator by 2 and cancel (x-1).) But why can't I cancel the (x-1) and the ((x/2)-0.5)?

    The limit about can be written as the (lim x->1 2x)*(lim x ->1 (x-1))/(lim x->1 (x/2)-0.5)
    The limits x-> 1 of (x-1) and ((x/2)-0.5) are the same, so why can't I cancel them like the (x-5) in the first example?

    Sorry if this is a stupid question, it's just that I'm trying to teach myself this and most material I can find doesn't discuss the ideas behind the math.

    Thanks :)
  2. jcsd
  3. Jul 12, 2012 #2


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    Hey Flumpster and welcome to the forums.

    You are correct about limits not necessarily being equal to the value of a function evaluated at that point.

    Basically what we say is that if the function equals the limit across an interval then the function is continuous. This means that we can take a pen and if we drawn the function, we can't ever take the pen off the paper to draw that function. If this is not the case (like your first example), then there will be 'holes' (like at when x = 5, the value of the function does not exist, and it is discontinuous at that point).

    The reason why limits are important is that in calculus, we can define using this definition if something is differentiable based on whether the derivative is 'continuous' (i.e. doesn't have the situation like with the division by zero like you had in the first case for the function), and this is used everywhere in the proof parts of calculus that the applied people take for granted.

    So yes, it's great that you picked up on that because a lot of people actually don't and get confused when they hear that limits are different from evaluating a function at that point.

    Also the other thing is that sometimes the limit doesn't exist and this is important as well but that's for another day.
  4. Jul 12, 2012 #3


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    A basic theorem on limits, that is unfortunately often over looked is this:
    If f(x)= g(x) for all x except x= a, then [itex]\lim_{x\to a}f(x)= \lim_{x\to a}g(x)[/itex].

    That is, the values of the two functions at x= 0 is completely irrelevant to the value of the limit.

    To see why that is true, look at the definition of "limit"
    [itex]\lim_{x\to a} f(x)= L[/itex] if and only if, given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex] then [itex]|f(x)- f(a)|< \epsilon[/itex].

    The key is that "[itex]0< |x- a|[/itex]". Unfortunately, that is also often neglected in class presentations- I've been guilty of that myself.

    Consider this: f(x)= 3x- 2 for all x. What is the limit as x goes to 1? g(x)= 3x- 2 for all x except 1 and g(1)= 0. What is the limit as x goes to 1? h(x)= 3x- 2 for all x except 1 and h(1)= 100000. What is the limit as x goes to 1?

    They are, of course, all "1". What matters for the 'limit as x goes to 1" are the values of the function near 1, not at 1.

    Unfortunately, the many of the elementary examples given beginning students have "nice" property that [itex]\lim_{x\to a} f(x)= f(a)[/itex] that students start to think of the limit as being "a fancy way to refer to the value of the function". The reason for that is that functions having that property are "nice" (they are called "continuous") so our ways of writing functions have grown up to make the formulas for such functions especially easy while to write functions that are not continuous we have to resort to "piecewise" definitions as I did above. That is not a fundamental "mathematical" property of our notation. In a very precise sense "almost all" functions are not continuous anywhere.
  5. Jul 12, 2012 #4
    Thank you both for such detailed replies! I understand some ideas much more now.

    But I still don't fully understand why I'm not allowed to cancel the two terms in the second limit.

    It's like in the first limit I gave, you can cancel the two (x-5)'s because:

    a) Even though they approach zero, they are not zero. (Now I understand this bit properly :D -thankyou!)

    b) both limits evaluate to the same value. (Limit A = Limit B = y)

    c) the whole limit can be re-written as lim(x+5)*(Limit A)/(Limit B)

    In the second example, it looks like I can cancel the (x/2 - 0.5) and (x-1) because in that limit a), b), c) are also met. But this gives an incorrect answer.

    What am I missing here?
  6. Jul 12, 2012 #5

    D H

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    Simple: x-1 is not equal to x/2-1/2. Instead, x-1=2(x/2-1/2). Use this and you now have equal terms in the numerator and denominator which do cancel.
  7. Jul 12, 2012 #6
    Ok, D H, you mean that because the expressions (x-1) and (x/2-1/2) are not equal, even if I prove that in the limit they are equal, I can't cancel them?

    (You say they aren't equal. Surely in the limit they are equal? You can evaluate each one separately and show that)

    Again, sorry if this is a stupid question, I just would have expected that if in the limit they were both equal, say L = a, you could cancel like in an basic equation where a variable has been substituted -

    y(x+5)/(2x) and you substitute x=5 then you get 10y/10 and cancel the tens.

    Edit: A more accurate example would be where you are separately given that 2x = 10 and x+5 = 10, and you directly substitute the 10.
    Last edited: Jul 12, 2012
  8. Jul 12, 2012 #7
    Clearly x/2 - 1/2 = (x-1)/2. It's not about the terms being equal at the point where the limit is to be evaluated. It's about their equality in a small neighborhood around the point of the limit.

    [tex]\lim_{x \to 1} \frac{x-1}{(x-1)/2} = \lim_{x \to 1} 2 \frac{x-1}{x-1}[/tex]

    It should be obvious that, as we get close to x=1, the value of this fraction is 2. In fact, the value of this fraction is 2 everywhere but x=1, but this is just a special case.

    You can cancel terms that are algebraically equal everywhere in a limit (not just at the point the limit is going to). If your expression doesn't give you something like 0/0 when you evaluate it at the point, you can take the real number answer that you get and not worry about it. I wouldn't call this canceling of terms--it's just arithmetic.
  9. Jul 12, 2012 #8


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    "in the limit they are equal" is, in my opinion, bad terminology. What you mean is that they have the same limit. But what is important here is NOT the limit but how fast each goes to that limit. As has already been discussed, we can cancel as long as x is NOT equal to, in this case, 1. As long as x is not 1, x- 1 and (x/2- 1/2) are NOT the same.

    I have no idea what you are talking about. "Substitute the 10" into what formula for what variable?
  10. Jul 12, 2012 #9
    OK, now I think I get it.

    If I have limit x-> A f(x), and limit x-> A g(x), it's not relevant that at A both limits have the same value, because the values of f(x) as the function approaches A are different from the values of g(x) as g(x) approaches A, and we're looking at values approaching A.

    Is that right?
  11. Jul 12, 2012 #10
    Yeah, basically. What you've run into is that the property


    breaks down if the limits on the right don't exist or if you get 0/0 or anything strange like that. Such things are called "indeterminate forms". So instead of taking the limits of the functions individually and then dividing, you have to cancel the common factor out first.
  12. Jul 12, 2012 #11
    About what HallsofIvy said about how fast the functions get to a point -

    If I take the two expressions we've been talking about, (x-1) and (x/2-1/2) that both equal 0 at x=1, and check their value close to 1, say at x = 1.001 I get:

    x-1 = 1.001 - 1 = 0.001.

    x/2 -1/2 = 1.001/2 -1/2 = 0.5005 - 0.5 = 0.0005.

    Well, both functions at being evaluated at an equal distance from x=1 (x=1.001), yet x/2 -1/2 is twice as close to the answer we would get at x=1 (0).

    Is this sort of what he meant, or have I made a total idiot of myself?
  13. Jul 12, 2012 #12
    Yes, this is precisely what's going on.
  14. Jul 12, 2012 #13
    Thank you everybody so much for all this help and the fantastic explanations! You guys really really helped me out a lot, and I understand things much better now.

    This made me think of a few other related issues -can I post them to this thread and continue the discussion or do I need to make a new thread?
  15. Jul 12, 2012 #14

    Ray Vickson

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    To understand the limit of f(x) = (x-5)(x+5)/(x-5) as x --> 5, we are really asking: what does f(x) get closer and closer and closer to as x gets closer and closer and closer to 5 (without neccessarily ever equalling 5)? Well, for any x different from 5, we have f(x) = x+5, because we can cancel out the (x-5 in the numerator and the denominator. So, f(x) gets closer and closer and closer to 5+5 = 10, so that is the limit.

    For f(x) = 2x(x-1)/[(x/2)-1/2], when x ≠ 1 we have f(x) = 2x(x-1)/[(1/2)(x-1)] = 4x, so the limit as x --> 1 is 4.

  16. Jul 13, 2012 #15
    Thanks, RGV :)

    Anyway, here goes with the related issues this made me think of:

    1. Is there any way to formally define or quantify "how fast" the limits approach a point? (I realize "how fast" is terribly inexact, I just don't know how to express it better). Does it depend on the epsilon-delta definition?

    There was another issue that interested me but I can't post it now. Again, thanks everybody for your help!
  17. Jul 13, 2012 #16

    Ray Vickson

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    Yes, there are "speed" measures for limits, etc., and they are often used in the analysis of algorithms, to help explain differences in efficiency, etc. Look, for example, at functions f(h) that have limit = 0 as h --> 0. "Linear" behaviour would be like that of f(h) = c*h, where f approaches 0 at the same rate as h (although scaled by the factor c). So, if you reduce h by a factor of 2 you reduce f by a factor of 2, etc. You can have faster behaviour, such as f(h) = c*h^2, where f --> 0 at the squared rate of h. In this case, if you reduce h by a factor of 2 you reduce f by a factor of 4; if you reduce h by a factor of 10 you reduce f by a factor 100, etc. Of course, there are higher speeds, such as h^n for n > 2. There are also some functions that do not fit nicely into such speed concepts, such as f(h) = 1/log(h) for h > 0. Here, f(h) --> 0 as h --> 0, but not by any power like h^k for fixed k > 0.


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