encontrar el siguiente límite: $\lim\limits_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}$

Question

encontrar el siguiente límite

$f(x)$ es diferenciable en a $x=1$ $f(1)>0$

$\lim\limits_{x\to 1}\left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}$

¿por qué puedo sustituir $x$$1$, y eso será el límite (desde $f$ es diferenciable, es también continua...)

Answer 1

$$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}= e^{\lim_{x \to 1}(\frac{f(x)}{f(1)}-1)\frac{1}{\ln x}} = e^{\frac{1}{f(1)}\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}\frac{x-1}{\ln x}}=e^{\frac{1}{f(1)}\cdot f'(1)\cdot1}=e^{\frac{f'(1)}{f(1)}}$$

Answer 2

$\newcommand{\+}{^{\daga}}% \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle}% \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \cima {= \cima \vphantom{\enorme}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\mitad}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\a la derecha\vert\,}% \newcommand{\cy}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$

1. \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}
2. \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}