Let $x=x_N\ldots x_0$, $y=y_M\ldots y_0$ be integers with base-$b$ digits $0 \le x_i,y_j < b$. Concatenate them via

$$x \otimes_b y\triangleq x_N\ldots x_0 y_M\ldots y_0=b^{len_b(y)}x+y=b^{1-\{\log_b{y}\}}xy+y$$

where $len_b(y)=M=\lfloor\log_b{y}\rfloor+1=\log_b{y}-\{\log_b{y}\}+1$, thus extending to all $b \ne 1$.

$\lim_{b \rightarrow 1} b^{1-\{\log_b{y}\}}=1$ when $\log(y) > 0$, so that $\lim_{b \rightarrow 1}x \otimes_b y = xy+y$.

**Q**: When $b=1$ (unary), concatenation should correspond to addition. Why is $\lim_{b\rightarrow 1} x\otimes_b y \ne x+y$?

I understand there are some difficulties in viewing unary as an honest base.