Weird calculus limit How to find the following limit? 
$$
\lim_{n \to \infty} \dfrac{ 5^\frac{1}{n!} - 4^\frac{1}{n!}  }{ 3^\frac{1}{n!} - 2^\frac{1}{n!}   }
$$
Edit done to the question.
Thank you!
 A: Let $x = \dfrac1{n!}$. We then have $$\lim_{n \to \infty} \dfrac{5^{1/n!} - 4^{1/n!}}{3^{1/n!} - 2^{1/n!}} = \lim_{x \to 0} \dfrac{5^x-4^x}{3^x-2^x}$$
Now $$\lim_{x \to 0}\dfrac{a^x - 1}x = \log(a) \tag{$\star$}$$
\begin{align}
\lim_{x \to 0} \dfrac{5^x-4^x}{3^x-2^x} & = \lim_{x \to 0} \dfrac{\dfrac{5^x-1}x-\dfrac{4^x-1}x}{\dfrac{3^x-1}x-\dfrac{2^x-1}x}\\
& = \dfrac{\lim_{x \to 0} \dfrac{5^x-1}x-\lim_{x \to 0} \dfrac{4^x-1}x}{\lim_{x \to 0} \dfrac{3^x-1}x-\lim_{x \to 0} \dfrac{2^x-1}x}\\
& = \dfrac{\log(5) - \log(4)}{\log(3) - \log(2)}\\
& = \dfrac{\log(5/4)}{\log(3/2)}
\end{align}
A: Hint. Substituting $x = \frac{1}{n!}$, you may consider the limit $$ \lim_{x\to 0} \frac{5^{x} - 4^{x}}{3^{x} - 2^{x}} $$ instead. Of course, the answer is $$\frac{\log(5/4)}{\log(3/2)}. $$
A: $$\frac{5^x-4^x}{3^x-2^x}$$
$$=\frac{4^x\left(\left(\frac54\right)^x-1\right)}{2^x\left(\left(\frac32\right)^x-1\right)}$$
$$=2^x\frac{\frac{ \left(\frac54\right)^x-1 }x}{\frac{\left( \frac32\right)^x-1 }x}$$
So, $$\lim_{x\to0}\frac{5^x-4^x}{3^x-2^x}$$
$$=\lim_{x\to0}2^x\frac{\frac{ \left(\frac54\right)^x-1 }x}{\frac{\left( \frac32\right)^x-1 }x}$$
$$=\lim_{x\to0}2^x\cdot\frac{\lim_{x\to0}\frac{ \left(\frac54\right)^x-1 }x}{\lim_{x\to0}\frac{\left( \frac32\right)^x-1 }x}$$
$$=\frac{\ln\frac54}{\ln\frac32}$$ using $\lim_{h\to0}\left(\frac{a^h-1}h\right)=\ln a$
