Evaluate limit of $\lim_{x\to 1} \frac{e^{(n!)^x}-e^{n!}}{x-1}$

Question

$$\lim_{x\to 1} \frac{e^{(n!)^x}-e^{n!}}{x-1}$$

Since it is $e^{(n!)^x}$ and not $e^{x×(n!)}$ so I can't using the general form for $\frac{e^x-1}{x}$ as $x\to 0$ could somebody help? It'd better if without using l'hopital rules

Answer 1

Let

$$f(x)=e^{(n!)^x}=e^{e^{x\log{n!}}}$$

The limit we’re looking at is the derivative at $x=1$ of $f$. So let’s find the derivative of $f=e^{g(x)}$

$$f’(x)=g’(x)e^{g(x)}=\log{n!}e^{x\log{n!}} e^{e^{x\log{n!}}}$$

Now plug $x=1$

$$f’(1)=n!\log{n!}e^{n!}$$

Answer 2

$y=x-1, n!=a$, $\lim\limits_{y\to 0}\dfrac{e^{a^{y+1}}-e^a}{y} =e^a\lim\limits_{y\to 0}\dfrac{e^{a^{y+1}-a}-1}{y} =e^a\lim\limits_{y\to 0}\dfrac{e^{a^{y+1}-a}-1}{a^{y+1}-a}\cdot \frac{{a^{y+1}-a}}{y}=e^a\cdot a\cdot\ln a$