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Justification for evaluation of limits?

  1. Feb 12, 2015 #1
    I am curious about the process of evaluating a limit. Firstly, I know that if a function ##f(x)## is continuous then one can usually just plug in the the number that ##x## is approaching in the limit, since criteria for a continuous function is that ##\lim_{x \to a}f(x) = f(a)##. However, what if we have the function ##f(x) = \frac{(x-1)^2}{(x+1)}##, then if we try to evaluate the limit ##\lim_{x \to -1}\frac{(x-1)^2}{(x+1)}##, we do so by cancelling the linear factors to get ##f(x) = x - 1##, which then gives ##\lim_{x \to -1}(x - 1) = -2##. But what is the justification that we're allowed to cancel out the factors when evaluating the limit? This produces an entirely new function, since ##f(x) = x - 1## is obviously different than ##f(x) = \frac{(x-1)^2}{(x+1)}##. Why are we allowed to manipulate one function into another in this way when evaluating limits? What guarantees that ##\lim_{x \to -1}(x - 1) = \lim_{x \to -1}\frac{(x-1)^2}{(x+1)}##?
     
  2. jcsd
  3. Feb 13, 2015 #2
    Both functions are exactly the same function written differently except for x=-1 which is when one of them becomes undefined.
    The algebraic manipulations you used to rewrite the initial function are perfectly valid ones for all x not equal to -1.
    But this is ok since when you evaluate a limit you want the value to which f(x) gets closer and closer, not the actual f(a).

    I guess to say it another way, you're replacing your old function with another function which is continuous at x=a, as oppossed to the original one, and which is also equal to the original except maybe at x=a, so that you can evaluate the limit by substitution. Both functions will have the same limit as x approaches a, if the limit exists, because both functions are identical everywhere except at x=a.

    Edit: I didnt see how you manipulated the function when i replied, what linear factor did you cancel?
    In any case, my argument remains unchanged except of course for the fact that i now think the algebraic manipulations of your example are not valid.
     
    Last edited: Feb 13, 2015
  4. Feb 13, 2015 #3

    Mark44

    Staff: Mentor

    No, this is incorrect. The limit here does not exist, because the left-side limit and the right-side limit are not equal.
     
  5. Feb 17, 2015 #4

    HallsofIvy

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    I presume the real difficulty you see is that the denominator goes to 0 at x= 1. There is no problem otherwise: 6/2 is equal to 3

    One very important, but often overlooked, property of limits is this: "If f(x)= g(x) for all x other than a, then [itex]\lim_{x\to a} f(x)= \lim_{x\to a} g(x)[/itex]".
    That follows from the definition of "limit":
    [itex]\lim_{x\to a} f(x)= L[/itex] if and only if, for any [itex]\epsilon> 0[/itex], there exist [itex]0< \delta[/itex] such that if [itex]0< |x- a|< \delta[/itex] then [itex]|f(x)- L|< \epsilon[/itex].

    Note that [tex]0< |x- a|[/tex]. What happens at x= a is irrelevant to the limit. In the case [tex]\frac{x^2- 1}{x- 1}[/tex] is equal to x+ 1 for all x except x= 1 but that is enough.
     
  6. Mar 3, 2015 #5
    draw a graph for both the initial and the final functions and see for yourself if they are different plus there is an error in your question.
     
  7. Mar 4, 2015 #6

    HallsofIvy

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    If you mean "draw graphs for both [itex]y= (x^2- 1)/(x- 1)[/itex] and [itex]y= x+ 1[/itex]", if you do it correctly you will see that they area NOT the same! The graph of [itex]y= x+ 1[/itex] is a straight line through (0, 1) with slope 1. The graph of [itex]y= (x^2- 1)/(x- 1)[/itex] is that same straight line except there is a "hole" at (1, 2).
     
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