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Question about limits

  1. Sep 14, 2014 #1
    Hello.

    Let us say that we have a graph f(x)=2x when x does not equal 3, and f(x)=undefined when x=3

    The limit of f(x) as x→3, is still 6.

    But what about the limit of 3+dx? Does it exist?

    I am not up to the derivatives section of my class, so maybe there is something I am not understanding yet.

    Because 3+dx is the smallest possible number after 3...and the limit of 3+dx=2(3+dx)=6+dx, if you just plug it in to the function. But, immediately to the left of 3+dx, f(x) is not defined, so the graph isn't getting closer and closer to 3+dx as you approach it from the left hand side...because the graph is actually getting closer and closer to being undefined.

    So would the limit not exist for 3+dx in this example?

    Thanks
     
  2. jcsd
  3. Sep 14, 2014 #2

    jedishrfu

    Staff: Mentor

    if you consider dx to be delta x then the limit at x+deltax = f(x+deltax)

    it doesn't matter how small delta x is it could be .1 .01 .001 ... or .000000000000000000000000000000000000000000001 and smaller
    and the limit is still by definition of the function f(x) still f(x)

    only when x + delta x = 3 ie delta x = 0 is the limt 6 but the function is undefined. It may boggle your mind in a physical sense but when doing math you live in a geometrical world where things can have zero size (a point) zero length or zero thickness (a plane)...
     
  4. Sep 14, 2014 #3
    But isn't DX defined to be infinitely small? So that nothing could be in between 3+DX and 3?


     
  5. Sep 14, 2014 #4
    No matter what non-zero value you choose for dx there is always a point 3 + dx/2 between 3 + dx and 3.
     
  6. Sep 14, 2014 #5

    True. Still, doesn't there have to be SOME number over there where there's no limit ? There has to be some endpoint before the function becomes undefined....no?
     
  7. Sep 14, 2014 #6
    There is no limit at x if the function jumps at x. So if you find that the "limit" of f(x) at x approaching from the left is different from approaching from the right, you say there is no limit at x.
     
  8. Sep 14, 2014 #7
    What number is it in the function I described where the function jumps?
     
  9. Sep 14, 2014 #8

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    No!!!!!!!!!!!!!!!!! As a previous post said, there is ALWAYS another number between two different real numbers.

    Unless you are working with hyperreal numbers (and from your OP I don't think you are), "3 + dx" is not a number, if "dx" is supposed to mean something related to "dy/dx". Vague ideas about "infinitesimally small numbers" will only confuse you. There is no such thing as "the smallest possible number after 3".

    If dx is an (ordinary) real number, if doesn't matter if it's 0.0001 or 10-1000000000000000000. There are an infinite number of different numbers between 3 and 3+10-1000000000000000000, just like there are an infinite number between 3 and 3.0001.

    But don't feel too bad about having problems with this. The ancient Greeks got tied in knots as well over the paradox of Achilles and the tortoise, etc.
     
  10. Sep 14, 2014 #9

    AlephZero

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    I doesn't jump at any number "in" the function, because you said the function was not defined for the number 3.
     
  11. Sep 14, 2014 #10
    I understand. That's crazy
     
  12. Sep 22, 2014 #11

    Stephen Tashi

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    Science Advisor

    No.

    In the standard texts on calculus "DX" by itself has no meaning. (In the somewhat eccentric approach of "nonstandard analysis", it might have.) The use of symbols such as "dy" and "dx" as if they were numbers is a handy way to remember certain rules in calculus. It is a mnemonic device. You can also made intuitive idea of some ideas in calculus if you think of "dy" and "dx" as numbers. However, thinking this way sometimes leads to wrong conclusions. There is no isubstitute for understanding the actual definitions of limit and derivative (the so-called epsilon-delta definitions).
     
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