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Need help understanding limits, and their evaluation

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    I don't know how to prove that the lim [itex]_{x-->a}[/itex] f(x) =f(a)

    2. Relevant equations
    basically how do we know that the limit of the function near a point a, is f(a). My textbook (Calculus Michael Spivak, 4th ed.) says that we prove this using theorem 2, which is the properties of limits...


    3. The attempt at a solution
    I really don't understand, the textbook says we can prove this directly, but it doesn't show it
     
  2. jcsd
  3. Oct 9, 2012 #2
    You need to have a particular function, not generic f(x), or at least you need to know some properties of the function to prove anything about its limits.
     
  4. Oct 9, 2012 #3

    HallsofIvy

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    Saying that "[itex]\lim_{x\to a} f(x)= f(a)[/itex]" is the definition of "continuous at x= a" and the great majority of functions are NOT continuous so you certainly cannot prove that for all f.

    How you would prove it for a specific function depends on the function, as well as what limit theorems you have learned.

    For example, to prove that f(x)= 3x- 4 is continuous for all x, that is, to prove that [itex]\lim_{x\to a} 3x- 4= 3a- 4[/itex], directly from the definition, I would look at the definition of "[itex]\lim_{x\to a} f(x)[/itex]" and see that it involves "[itex]|f(x)- L|< \epsilon[/itex]". Here, f(x)= 3x- 4 and L, the limit I want to prove, is f(a)= 3a- 4. So I know that I need to look at |(3x-4)- (3a- 4)|= |3x- 3a|= 3|x- a|. And I need to prove that [itex]3|x- a|< \epsilon[/itex] as long as [itex]|x- a|< \delta[/itex] where I need to be able to choose a suitable [itex]\delta[/itex] for any given [itex]\epsilon[/itex]. Comparing "[itex]3|x-a|< \epsilon[/itex]" and "[itex]|x- a|< \delta[/itex]" it immediately occurs to me that I need to choose [itex]\delta= \epsilon/3[/itex] (or, strictly speaking any smaller number).

    But if I have the "limit theorems" that
    1) [itex]\lim_{x\to a} C= C[/itex] for any constant C
    2) [itex]\lim_{x\to a} x= a[/itex]
    (those are often called the "trivial limits")
    3) if [itex]\lim_{x\to a} f(x)= F[/itex] and [itex]]\lim_{x\to a} g(x)= G[/itex], then [itex]\lim_{x\to a} (f+ g)(x)= F+ G[/itex]
    4) if [itex]\lim_{x\to a} f(x)= F[/itex] and [itex]]\lim_{x\to a} g(x)= G[/itex], then [itex]\lim_{x\to a} (fg)(x)= FG[/itex]
    5) if [itex]\lim_{x\to a} f(x)= F[/itex] and [itex]]\lim_{x\to a} g(x)= G[/itex], AND [itex]G\ne 0[/itex], then [itex]\lim_{x\to a} (f/g)(x)= F/G[/itex]

    I can immediately say that [itex]\lim_{x\to a} 3x- 4= (lim_{x\to a}3)(\lim_{x\to a}x)+ (\lim_{x\to a} -4)= 3a- 4[/itex]

    If I also know the very important but often overlooked limit theorem
    6) If f(x)= g(x) for all x close to but NOT equal to a, then [itex]\lim_{x\to a}f(x)= \lim_{x\to a} g(x)[/itex]

    I can argue that, for x not equal to a, [itex](x^2- a^2)(x- a)= x+ a[/itex] so that [itex]\lim_{x\to a}(x^2- a^2)/(x- a)= \lim_{x\to a} x+ a= 2a[/itex]. (Which I could NOT do using (5) because the denominator goes to 0.)
    So that I have proved that the function "[itex]f(x)= (x^2- a^2)/(x- a)[/itex] as long as [itex]x\ne a[/tex] and f(a)= 2a" is continuous at x= a.
     
    Last edited: Oct 9, 2012
  5. Oct 9, 2012 #4
    Thank you for the clear explanation
     
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