Why is $\lim_{b \rightarrow 1} concat_b(x,y) \ne x+y$ for base-$b$?

(Too long for a comment.)

$x \otimes_b y\triangleq x_N\ldots x_0 y_M\ldots y_0=b^{\text{len}(y)}x+y$

This works for positional numerals in base $b \ne 1$, and it works because 1) multiplication by $b$ is equivalent to a left-shift of all digits with a $0$ inserted at the rightmost position, and 2) $\text{len}(y)=\lfloor\log_b{y}\rfloor+1$ represents the number of digits in the representation of $y$.

But in the unary numeral system 1) there is no digit $0$ and no notion of left-shifting the digits with "nothing" inserted at the right, and 2) the length of the representation is simply $\text{len}(y)=y$. Because of that, it is not possible to define concatenation for unary numerals as a limit case of concatenation for positional numerals of base $b \ne 1$. Instead, concatenation for unary numerals is simply $\,x \otimes_1 y = x + y\,$.