find the following limit

$f(x)$ is differentiable at $x=1$ and $f(1)>0$

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

why can i just substitute $x$ for $1$ and thats will be the limit (since $f$ is differentiable, it is also continuous...)

  • $\begingroup$ If you substitute $x = 1$, your exponent is poorly behaved.... $\endgroup$ – user61527 Dec 29 '13 at 21:14
  • $\begingroup$ ooh..., so what do i need to do? $\endgroup$ – guynaa Dec 29 '13 at 21:15
  • $\begingroup$ Hint: $\log f$ is also differentiable at $1$. $\endgroup$ – Julien Dec 29 '13 at 21:18
  • $\begingroup$ possible duplicate of Limits and derivatives - two questions $\endgroup$ – GinKin Dec 30 '13 at 16:27

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

  • 1
    $\begingroup$ why is the first step allowed? if you want to use $e$ shouldn't the power be surrounded with $\ln$? $\endgroup$ – guynaa Dec 29 '13 at 21:33
  • $\begingroup$ @guynaa $a^b=e^{b \ln a}$ $\endgroup$ – Mikulas Dite Dec 29 '13 at 21:48
  • $\begingroup$ but you dident put $\left(\dfrac{f(x)}{f(1)}\right)$ inside $\ln$... $\endgroup$ – guynaa Dec 29 '13 at 21:53

We have by the definition of the derivative: $$\lim_{x \to 1} \left(\dfrac{f(x)}{f(1)}\right)^{\frac{1}{\log(x)}}=\lim_{x \to 1}\exp\left(\frac{\log f(x)-\log f(1)}{x-1}\frac{x-1}{\log(x)}\right)\\=\exp\left(\frac{(\log f(x))'\big|_{x=1}}{(\log x)'\big|_{x=1}}\right)=\exp\left(\frac{f'(1)}{f(1)}\right)$$

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    $\begingroup$ could you please explain what you did in each step? $\endgroup$ – guynaa Dec 29 '13 at 21:29
  • $\begingroup$ In the first step I just rewrite the expression using $\exp\log$ functions and to appear the form $$\frac{f(x)-f(a)}{x-a}$$ and the second step is to pass to the limit $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$$ $\endgroup$ – user63181 Dec 29 '13 at 21:33
  • $\begingroup$ @guynaa. If $\lim^|_{x→α}f(x)=1$ and $\lim_{x→α}g(x)=∞$ then there is "the shortcut"$\lim_{x→α}(f(x)^{g(x)}=e^{\lim_{x→α}(f(x)−1)g(x)}$ $\endgroup$ – medicu Dec 29 '13 at 21:37
  • $\begingroup$ can you point me to a reference to this "shortcut"? $\endgroup$ – guynaa Dec 29 '13 at 21:41
  • $\begingroup$ It is a consequence of remarkable limit $\lim_{n \to \infty} (1+ \frac{1}{n})^n=e$ $\endgroup$ – medicu Dec 29 '13 at 21:45

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

  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}

Let $L$ be the limit you want to find. Then using L'Hospital's rule,

\begin{align*} \ln L &= \lim_{x \to 1} \ln \left(\frac{f(x)}{f(1)}\right)^{1/\ln{x}} \\ &= \lim_{x \to 1} \dfrac{\ln f(x) - \ln f(1)}{\ln x} \\ &= \lim_{x \to 1} \dfrac{\dfrac{f'(x)}{f(x)}}{\dfrac 1 x} \\ &= \lim_{x \to 1} \frac{x f'(x)}{f(x)} \end{align*}

Can you fill in the details, and find the limit from here?

  • $\begingroup$ i cant use L'Hospital's rule, we haven't covered it yet $\endgroup$ – guynaa Dec 29 '13 at 21:20

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