Find the following limit: $\lim\limits_{x \to 1} \left(\frac{f(x)}{f(1)}\right)^{1/\log(x)}$
We have by the definition of the derivative: $$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}=\lim_{x \to 1}\exp\left(\frac{\log f(x)-\log f(1)}{x-1}\frac{x-1}{\log(x)}\right)\\=\exp\left(\frac{(\log f(x))'\big|_{x=1}}{(\log x)'\big|_{x=1}}\right)=\exp\left(\frac{f'(1)}{f(1)}\right)$$
$$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}= e^{\lim_{x \to 1}(\frac{f(x)}{f(1)}-1)\frac{1}{\ln x}} = e^{\frac{1}{f(1)}\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}\frac{x-1}{\ln x}}=e^{\frac{1}{f(1)}\cdot f'(1)\cdot1}=e^{\frac{f'(1)}{f(1)}}$$
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- \begin{align}\color{#0000ff}{\large% \lim_{x \to 1}\bracks{\fermi\pars{x} \over \fermi\pars{1}}^{1/\ln\pars{x}}} &= \lim_{x \to \infty} \bracks{\fermi\pars{1 + 1/x} \over \fermi\pars{1}}^{1/\ln\pars{1 + 1/x}} = \lim_{x \to \infty} \bracks{1 + {\fermi'\pars{1} \over \fermi\pars{1}}\,{1 \over x}}^{x} \\[3mm]&= \lim_{x \to \infty} \exp\pars{x\ln\pars{1 + {\fermi'\pars{1} \over \fermi\pars{1}}\,{1 \over x}}} = \lim_{x \to \infty} \exp\pars{x\bracks{{\fermi'\pars{1} \over \fermi\pars{1}}\,{1 \over x}}} \\[3mm]&= \color{#0000ff}{\large\expo{\fermi'\pars{1}/\fermi\pars{1}}} \end{align}
- \begin{align} \lim_{x \to 1}\ln\pars{\bracks{\fermi\pars{x} \over \fermi\pars{1}}^{1/\ln\pars{x}}} &= \lim_{x \to 1}{\ln\pars{\fermi\pars{x}/\fermi\pars{1}} \over \ln\pars{x}} = \lim_{x \to 1}{\fermi'\pars{x}/\fermi\pars{x} \over 1/x} = {\fermi'\pars{1} \over \fermi\pars{1}} \\[3mm]&\quad\imp\quad \color{#0000ff}{\large% \lim_{x \to 1}\bracks{\fermi\pars{x} \over \fermi\pars{1}}^{1/\ln\pars{x}}} = \color{#0000ff}{\large\expo{\fermi'\pars{1}/\fermi\pars{1}}} \end{align}