Question

Can anyone explain it?
We say that f(x) approaches infinity as x approaches to \(x_0\), and write \(lim_{x \rightarrow x_0} f(x) = \infty\), if for every positive real number B there exists a corresponding \(\delta\) >0 such that for all x, 0<| \(x-x_0\) |< \(\delta\) => f(x) > B

Answer 1

Actually, I don't understand the part ''if for every positive real number B.......... => f(x) > B''

Answer 2

In words you could say that if the limit exist, no matter how big a number B you pick can you can always find a region sufficiently close to \(x_0\) so that all \(x\) within that region makes \(f(x)\) bigger than B.
Did that make it clearer?

Answer 3

If not, maybe an example of this will really get the point home. So if I claim the following limit exists,
\[ \lim_{x\rightarrow0}\frac{1}{x}=\infty \]
I basically say that you can throw any number at me, no matter how big and I can always find a \(\delta\) such that the function \(f(x)=\frac{1}{x}\) always is bigger than your number within the region \(0<|x-x_0|<\delta \).

So lets say you pick 1 000 000, then I choose \(\delta=0.000001\) so that if you evaluate \(\frac{1}{x}\) for any number between 0 and \(\delta\), it will be bigger than your 1 000 000.
This doesn't prove my claim, to prove it I have to show that you can pick any positive number, not just 1 million. But it illustrates what it means to say something has an infinite limit.

Answer 4

Thanks! But must delta be corresponding to the number we choose? In your case, \(x_0\) = 1000000, and f(1000000) = 1/1000000 = 0.000001.
So, when I choose 5x10^7, delta must be 2x10^(-8), right?

Answer 5

Looks like I didn't get it...

Answer 6

Well, most often in practice we would never use actual numbers, this was just a illustration of what the definition says. But it's nothing wrong with your example. The point is that you can always find the delta, no matter what number we choose.
Also in my case \(x_0=0\) and not 1 million, since it's 0 we are approaching. Maybe a picture will clear everything up: