Let $P(n) = a^{P(n-1)}-1$ such that for all $n = 2 ,3 ,$ and so on. And let $P(1) = a^x -1$ where a belongs to all real positive numbers, then we have to evaluate $\lim\limits_{x\to0} P(n)/ x$ where $x$ apoaches to zero .
I tried it and found that there is indeterminancy of $0/0$
But I cannot go further as I cannot apply series and l hospital.
Let $f(x) = a^x - 1$, your $P(n)$ is really a function in $x$ obtained by composition of $f$ with itself for $n$ times. $$P(n) = f^{\circ n}(x) \stackrel{def}{=} \underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}}(x)$$ Since $f$ is a differentiable function in $x$ and $f(0) = 0$, you have $f^{\circ n}(0) = 0$ and
$$\ell_n \stackrel{def}{=}\lim_{x\to 0}\frac{P(n)}{x} = \lim_{x\to 0}\frac{f^{\circ n}(x)}{x} = \lim_{x\to 0}\frac{f^{\circ n}(x) - f^{\circ n}(0)}{x-0} = \left.\frac{d}{dx} f^{\circ n}(x)\right|_{x=0} $$ Apply chain rule to $f^{\circ k}(x)$ for a generic positive integer $k$, we have $$\frac{d}{dx}\left[ f^{\circ k} (x) \right] = f'(f^{\circ k-1}(x))\frac{d}{dx}\left[ f^{\circ k-1}(x)\right]$$ At $x = 0$, this give us $\ell_k = f'(0)\ell_{k-1}$. From this, we can deduce $$\lim_{x\to 0}\frac{P(n)}{x} = \ell_n = f'(0)\ell_{n-1} = f'(0)^2 \ell_{n-2} = \cdots = f'(0)^n = (\log a)^n$$