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Limits and 0/0

  1. Sep 4, 2003 #1
    Greetings.
    Let me first show how my teacher solves a limit.
    For example, take :
    Lim(x->6) (x+5)/(x-3)
    He takes the number that x is heading to (forgive me for the terminology), which is 6 in our case.
    Now put 6 in the place of x in the function after the limit, you get :
    6+5/(6-3) = 11/3
    Now, there are 4 conditions for the ratio that u will get :
    A: NonZero/NonZero : in this case there is no problem
    B: Zero/NonZero : no problem in this case too
    C: NonZero/Zero : in this case the limit will either be +[oo] or -[oo]
    D: Zero/Zero : in this case we have to use some tricks to make the limit back to the case of A, B or C.
    Everything i just said was what my teachers says.

    Today in class i came up with a question that (the way i see it) does not work on the rules of my teacher. Here it is :
    Define :
    Code (Text):

    f(x) = 2 , (x=1)
           x , (x[x=]1)
     
    Now, try to find the limit :
    Lim(x->1) (f(x)-1)/(x-1)
    If you use my teacher's way, you will reach 1/0, which means that the limit either equals +[oo] or -[oo]. But if you solve the limit, you will find that it equals 1 (unless i am missing sth).
    When i told my teacher about this, he said : "no no no no no (maybe 30 times), it is impossible", i told him "just see this question, and either tell me where i went wrong, or ur rule isn't right", he said "this is not MY rule" (then in a silly way) "if you want try to invent ur own theory then call it on your name", i tried to explain my way of solving the problem, but he wouldn't hear me.
    Anyway, here is my way :
    Take the limit of the numerator, and the limit of the denominator, then write both limits as a ratio, then apply my teacher's rules (A,B,C and D).

    Please tell me what do u think. Feel free to ask any questions.
    Thanks.
     
    Last edited: Sep 4, 2003
  2. jcsd
  3. Sep 4, 2003 #2

    chroot

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    What you've created is a point discontinuity. There is no limit defined at x=1.

    - Warren
     
  4. Sep 4, 2003 #3

    HallsofIvy

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    The limit lim(x->a) f(x) is the single number, if there is such, that f(x) nears as x nears a (without actually being equal to a).

    Your teacher's method works for continuous functions (indeed, "continuous functions" are DEFINED
     
  5. Sep 4, 2003 #4

    Hurkyl

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    You are neglecting a crucial condition for applying this "plug-in" rule; both the numerator and denominator must be continuous.

    The relevant theorem, taken directly from my advanced calculus text (but notation changed to suit this medium), says:

    Assuming that f and g are each defined on a deleted neighborhood of x = b and that f(x)→A as x→b
    and g(x)→B as x→b, then it is true that

    ...

    and if B[x=]0

    f(x)/g(x)→A/B as x→b


    Because of the way you defined your numerator, the hypotheses of this theorem does not hold.
     
  6. Sep 4, 2003 #5

    HallsofIvy

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    The limit lim(x->a) f(x) is the single number, if there is such, that f(x) nears as x nears a (without actually being equal to a).

    Your teacher's method works for continuous functions (indeed, "continuous functions" are DEFINED as functions for which this works!) and it can be show that functions made by the usual operations "add, subtract, multiply, divide, powers" are continuous as long as we do reach and "impossible" operation (divide by 0, take square root of a negative number, etc.)

    The function you give is not continuous (your f(x) is not continuous because the limit at x= 1 is clearly 2 (No matter how close to 1 x is, as long as x is not 1, f(x)= 2 so it's "close to" 2) but the value of the function is 1, not 2: the limit is not the value of the function- that itself "blows" your teacher's method, you don't need the "-1/(x-1)" part!). What you can do is take the "one-sided" limits. If we look only at x< 1, we can replace f(x) by 2:
    f(x)-1= 2-1= 1 so (f(x)-1)/(x-1)= 1/(x-1). That reduces to the "Nonzero/zero" case. Since x-1<1 here, the limit is "- infinity" (by the way, a lot of people, myself included, who say that saying the "limit is -infinity (or +infinity)" is just another way of saying the limit does not exist). If you take x> 1, you have f(x)-1= 2-1= 1 again, but now x-1> 0. The limit is "+ infinity". Those results tell us that this function is not getting "close to" any specific number and there is no limit.

    (I'm impressed that you are thinking of "piecewise" functions. Most beginning students shun those like the plague.)
     
  7. Sep 5, 2003 #6
    Thanks all.
    chroot said
    Can you explain why please ?
    It is possible to have a function that has a discountinuity at a certain point say x=b, and still has a limit when x gets near b, my function is one of them (as far as i see). Please read to the end of the reply to see why.
    Hurkyl said
    Well, i think the theory that the calculus book is talking about is only the cases A and B of my teacher's idea, since (as you can see), my teacher's method has the denominator=0 sometimes (but will not accept the answer directly).
    I understand that the plug-in rule is not acceptable when the limit of the denominator is 0, but my teacher does not look at the matter this way, my teacher plugs in the values and uses the results to decide how to deal with the limit (as you can see from my first post)
    HallsofIvy
    After reading ur post i felt a little bit lost, let me explain why.
    I am ok with the first two paragraphs, but the trouble comes from the third paragraph.
    I see the exact opposite, the value of the function f(x) at x=1 is 2=f(1), and the limit at x=1 is clearly 1 !!
    This is what i actually did before writting the first post, and here is my method.
    First, i redefined the original function in a more elegant way
    Code (Text):

    f(x) = x , x < 1
           2 , x = 1
           x , x > 1
     
    Then, i took the right handed limit :
    lim(x->1+) (f(x)-1)/(x-1) = lim(x->1+) (x-1)/(x-1) = lim(x->1+) 1 = 1
    (since when x gets near 1 from the right, it will be bigger than 1, then f(x) (in this case) will be x)
    Doing the same to the left handed limit, you get :
    lim(x->1-) (f(x)-1)/(x-1) = 1
    So, we conclude :
    lim(x->1) (f(x)-1)/(x-1) = 1

    Am i right ?
    Thank you .

    (Thanks in advance for your following replies).
    (Edited for a missing symbol)
     
    Last edited: Sep 5, 2003
  8. Sep 5, 2003 #7

    HallsofIvy

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    You are right, I misread the function.

    [QU0TE]This is what i actually did before writting the first post, and here is my method.
    First, i redefined the original function in a more elegant way

    code:--------------------------------------------------------------------------------
    f(x) = x , x < 1
    2 , x = 1
    x , x > 1
    --------------------------------------------------------------------------------

    Then, i took the right handed limit :
    lim(x->1+) (f(x)-1)/(x-1) = lim(x->1+) (x-1)/(x-1) = lim(x->1+) 1 = 1
    (since when x gets near 1 from the right, it will be bigger than 1, then f(x) (in this case) will be x)
    Doing the same to the left handed limit, you get :
    lim(x->1-) (f(x)-1)/(x-1) = 1
    So, we conclude :
    lim(x->1) (f(x)-1)/(x-1) = 1

    Am i right ?[/QUOTE]

    Yes, with this new function the limit is 1.

    It appears to me that your teacher DOES "look at the matter this way". Your teacher is saying that the "plug in rule" GIVES the right limit if the denominator is not 0. If the denominator is 0, then then you look at whether or not the numerator is 0. If not, the limit does not exist. If yes, then you need to look more closely (my preferred way of saying "use tricks"!).

    By the way, the "plug in rule" does NOT say "if g(a) is not 0 then lim f(x)/g(x)= f(a)/g(a)". This is only true if f and g are themselves continuous at a.

    What you are really doing is using the limit theorems:
    If lim f(x)= A, lim g(x)= B then
    a) lim(f(x)+ g(x))= A+ B
    b) lim(f(x)- g(x))= A- B
    c) lim(f(x)*g(x))= A*B
    d) lim(f(x)/g(x))= A/B as long as B is not 0 (all limits as x->a)

    together with the fact that lim(x->a) x= a and lim(x->a) c= c
    (c a constant). From those one can show that all polynomials are continuous (lim f(x)= f(a)) for all a and rational functions are continuous at a as long as a does not make the denominator 0.

    Apparently you have not been introduced to continuous functions yet.

    "Plug in" only works for continuous functions. Actually, "almost all" functions are NOT continous anywhere but continuous functions are so nice that our function notation has developed so that practically every function we can write easily is continuous.
     
  9. Sep 5, 2003 #8
    this is not true.
     
  10. Sep 5, 2003 #9

    chroot

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    lethe,

    Yeah, yeah, I know, I read the damn function wrong too. The limit's one.

    - Warren
     
  11. Sep 5, 2003 #10
    Exactly, this is what my teacher sees, and this is what i am trying to proove wrong using my problem :smile:.
    As you can see, when you 'plug in' the value, you will have 0 in the denominator, and if you look at the numerator its value will not be zero, and therefore (according to my teacher's idea that u are perfectly expressing when you said "If not, the limit does not exist.") the limit does not exist, althought if we investigate it more we will find it exists.
    Well, actually, i know what they are pretty well. The point is that my teacher isn't really accurate, actually ... isn't accurate at all! He told us about the plug in rule without even mentioning that it only works under certain conditions.

    What i meant by 'using tricks' is using mathematical methods to change the limit which's plug in is 0/0 to a Nonzero/Nonzero value.

    The real question lies here.
    My teacher thinks we should start 'using tricks' when the result of plugging in the value of x in the limit is 0/0 (please not that he DOES NOT investigate wether or not the function is continuous).
    What i am suggesting is that we should start 'using tricks' when the ratio of the limit of the numerator divided by the limit of the denominator is 0/0. This way wether the function is continous or not we will know we need to 'use the tricks'.
    Tricks include : Analyzing the numerator to a simpler form, multiplying both numerator and denominator by certain roots, adding and subtracting certain numbers/functions to the numerator ... etc.

    My question is only wether or not what i am suggesting is right.
    Thanks.

    EDIT :
    This is the problem, my teacher does NOT investigate about the functions' continuity before using the plug in rule.
     
    Last edited: Sep 5, 2003
  12. Sep 6, 2003 #11
    your teacher's methods work for continuous functions, of just plugging it in and seeing what you get. you just have a discontinuity but the limit IS defined and as you said, it is 1. to actually prove that it is one is not that hard, but requires some epsilon-delta stuff i won't go into here. maybe you should be one of the few who starts with advanced calculus rather than calculus cuz that's the box you're bumping up against.

    may your journey be graceful,
    phoenix
     
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