I have made two formulae. One tending to $e$ and one tending to $3$:
$$e = \lim_{n \rightarrow \infty} \left(\frac{2 \times 10^n - 1}{2 \times 10^n - 3}\right)^{10^n - 1}$$
$$3 = \lim_{n \rightarrow (-\infty)} \left(\frac{2 \times 10^n - 1}{2 \times 10^n - 3}\right)^{10^n - 1}$$
Is there a way of proving that these formulae are indeed true? Thanks.
The second one is true because $\lim_{n\to(-\infty)}10^n=0$, so we get $$\lim_{n\to(-\infty)}\bigg(\frac{2\times10^n-1}{2\times10^n-3}\bigg)^{10^n-1}=\bigg(\frac{-1}{-3}\bigg)^{-1}=3.$$ For the first one, note that $$\lim_{m\to\infty}\bigg(\frac{2m+1}{2m-1}\bigg)^m=\lim_{m\to\infty}\bigg(1+\frac{1}{m-1/2}\bigg)^{m-1/2}\bigg(1+\frac{1}{m-1/2}\bigg)^{1/2}.$$ The second bracket goes to $1$ and the first to $e$. Since your sequence is just a subsequence of this one, it has the same limit.