You're just using circular logic here anyway. You haven't proved that lim n->inf 1 / 10n is 0. What if it's a really really small number that's immediately above 0?

Delta-epsilon proof, if you really need it.

Sure, throw it my way. I want to see how you get the above result without assuming exactly what we are in the middle of debating.

If we define real numbers as any decimal expansion, then this wouldn't hold.

OK, if this really is a problem for you.

A limit of a sequence is defined as follows.

*L is the limit of a sequence, x*_{n}, if and only if for every ε > 0, there exists an N such that for any positive integer n greater than N |x_{n}-L|< ε. Setting L = 0 and x

_{n} = 1/10

^{n}, we obtain the following statement to that we need to prove.

Proposition: For any ε > 0, there exists an N such that for any positive integer n greater than N |1/10

^{n}|< ε.

Proof: Let N = -log

_{10} ε. This defines an N for any ε > 0. It can be shown that |1/10

^{n}|< ε for all n > N.

It is given that n > N. Subtituting N = -log

_{10} ε, we have

n > -log_{10} ε.

We can solve for ε, to find

1/10^{n} < ε.

Taking the absolute value, we have

|1/10^{n}| < ε

since |ε|=ε because ε > 0.

Thus, we have shown for any ε > 0 there exists an N such that for any n > N, |1/10

^{n}| < ε namely N = -log

_{10} ε. The proposition is proven and the limit is 0.

Can we please let this die?

No.