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Two limits

  1. Apr 21, 2013 #1
    I have the expression:

    limb->-8[½b^2]-limb->8[½b^2]

    Is it rigorously defined how to calculate this? The question arose because I want the additivity of improper integrals to work and if you take the integral of x from minus infinity to infinity to work the expression above must be zero.
     
  2. jcsd
  3. Apr 21, 2013 #2
    There are no domain restrictions that we need to be aware of here, so we can just say that ##\displaystyle \lim_{b\rightarrow -8}\frac{b^2}{2} - \lim_{b\rightarrow 8}\frac{b^2}{2} = \frac{(-8)^2}{2}-\frac{8^2}{2} = 0##

    If you want rigor, consider the ##\epsilon ,\delta## definition of a limit:

    $$\lim_{x\rightarrow \alpha}f(x) = \mathfrak{L} \iff \forall\epsilon>0 \ \exists\delta>0:\forall x (0<|x-\alpha|<\delta \Rightarrow 0<|f(x)-\mathfrak{L}|<\epsilon)$$
     
  4. Apr 21, 2013 #3

    Bacle2

    User Avatar
    Science Advisor

    In other words, both limits exist, by continuity of f(x)=x^2 , so you can calculate them individually and subtract. Remember that continuity is equivalent to lim_x-->xo f(x)= f(xo) , so if you accept continuity of
    f(x)=x^2 ( or prove it in the way Mandelbroth suggested ), the result follows.
     
  5. Apr 21, 2013 #4
    lmao. I now see that what appeared as the number 8 was supposed to be infinity. I think that will change the answer from you a bit .
     
  6. Apr 22, 2013 #5

    Mark44

    Staff: Mentor

    You can do this:
    $$ \int_{-\infty}^{\infty} x~dx = \lim_{b \to \infty} \int_{-b}^b x~dx$$
    $$ = \lim_{b \to \infty} \left.(1/2)x^2 \right|_{-b}^b$$
    Can you continue from here?
     
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