But How Many Digits Are There?

We note that $\begin{cases} x_n = 1+10^{-(n+1)} \\ y_n = 1 - 10^{-(n+1)} \end{cases}$ . Let $a = 10^{-(n+1)}$ $\implies \begin{cases} x_n = 1+a \\ y_n = 1 - a \end{cases}$ and as $n \to \infty$ , $a \to 0$ .

Then, we have:

$\begin{aligned} L & = \lim_{n \to \infty} 10^{n+1} \left(x_n^{x_n} - y_n^{y_n} \right) \\ & = \lim_{a \to 0} \frac 1a \left((1+a)^{1+a}-(1-a)^{1-a} \right) \quad \quad \small \color{#3D99F6}{\text{Since this is a }0/0 \text{ case, we can use L'Hôpital's rule.}} \\ & = \lim_{a \to 0} \frac 11 \left((1+a)^{1+a} (\ln(1+a)+1)-(1-a)^{-a}(a-1)(\ln(1-a)+1) \right) \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down w.r.t }a.} \\ & = \boxed{2} \end{aligned}$

Writing

$x_{n} = 1+ 10^{-(n+1)}$

And

$y_{n}= 1-10^{-(n+1)}$ .

Now we know for small $\kappa$ , $\forall n \in R$

$(1+\kappa )^{n} = 1+ n\kappa$

Using the above approximation , value of limit is

$2$