We know, $6^n \equiv 6(mod \ 10) \rightarrow [1]$
Let, $M=\lim_{n \to \infty} (6^n)$ then, $M$ is tending to $\infty$ and, from $[1]$ we have $6$ as last digit of $M$ so, have we predicted the last digit of $\infty$ or, there is some flaw in above manipulations?
On
Infinity is NOT a number!
My teacher showed me a nice way of thinking about infinity in context of limits
Let's say, that you are throwing a rock. The limit is a place, where the rock lands. If you throw a rock strong enough, it will go to the outer space and will never land on the ground. It will 'land' in the space, so the space is it's limit. Infinity is like space - it is container, not a point on ground (number)
Also the 'last digit of infinity' (even 'last $N$ digits of infinity') can be any number ;): $$\forall k \in (0, 10^N -1) \lim 10^n+k = \infty$$
You are implicitly admitting that
$$\left(\lim_{n\to\infty}6^n\right)\bmod10=\lim_{n\to\infty}(6^n\bmod 10)$$
and conclude that both expressions equal $6$. This is the "forbidden step": the first limit doesn't exist and you can't take its modulo.
If you bypass this logics, you can similarly show that the last digit is $0$ or $5$, a perfect contradiction.
This paradox arises because you are going to infinity using a sequence only made of numbers that end in $6$ (you could as well have chosen the sequence $10n+6$). But this sequence is lacking many numbers, such as those ending in $7$ (i.e. $10n+7$) also forming sequences that tend to infinity. So the limit below does not exist: $$\lim_{n\to\infty}(n\bmod10)$$ and infinity has no last digit.