Two limits

1. Apr 21, 2013

aaaa202

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. Apr 21, 2013

Mandelbroth

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:

$$\lim_{x\rightarrow \alpha}f(x) = \mathfrak{L} \iff \forall\epsilon>0 \ \exists\delta>0:\forall x (0<|x-\alpha|<\delta \Rightarrow 0<|f(x)-\mathfrak{L}|<\epsilon)$$

3. Apr 21, 2013

Bacle2

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.

4. Apr 21, 2013

aaaa202

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 .

5. Apr 22, 2013

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?