finding a limit of a general function. ${\rm f}$ is differentiable at ${\rm f}\left(1\right),\,$
${\rm f}\left(1\right) > 0$.
Calculate the following limit, and here's what I did:
$$
\lim_{n \to \infty }\left[%
{\rm f}\left(1 + 1/n\right) \over {\rm f}\left(1\right)\right]^{1/n}
=
\left[{\rm f}\left(1\right) \over {\rm f}\left(1\right)\right]^{0} = 1
$$
Now, On the one hand, I can't see why the above isn't true. On the other hand, I got this hunch something is wrong here.
Can you direct me please?
 A: What you did is something like the following.
$$\lim_{n\to\infty}\left(1+\frac 1n\right)^n=\left(1+0\right)^{\infty}=1.$$
As you know the correct answer is $e$.
The reason why you cannot do like this is that the exponent $n$ does influence $1+1/n$. 
Your case is the same as above. The exponent $1/n$ does influence the inside. So you cannot use limit to separate $n$s. 
Don't confuse this with the following :
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\cdot \left(1+\frac{1}{n}\right)=(1+0)\cdot (1+0)=1.$$
This is correct.  
A: I think you meant to write $$ \lim_{n \to \infty} \left( \frac{f(1 + \frac{1}{n})}{f(1)} \right)^{\!n},$$ not $1/n$, since the expression as written is trivially always $1$ given that $0 < f(1) < \infty$.
Consider the logarithm:  $$ \log \left( \frac{f(1 + \frac{1}{n})}{f(1)} \right)^{\!n} = \frac{\log f(1 + \frac{1}{n}) - \log f(1)}{1/n}. $$  Then by the definition of derivative $$ g'(a) = \lim_{t \to a} \frac{g(t) - g(a)}{t-a}$$ with the choice $a = 1$, $g = \log f$, and $t = 1 + 1/n$, we see that the limit of the logarithm is simply the derivative of $\log f$ at $x = 1$; i.e., $$\lim_{n \to \infty} \left( \frac{f(1 + \frac{1}{n})}{f(1)} \right)^n = \exp \left( \frac{d}{dx} \left[ \log f(x) \right]_{x=1} \right) = \exp \frac{f'(1)}{f(1)},$$ which is the more interesting case.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
When $n \gg 1$:
$$
\bracks{\fermi\pars{1 + 1/n} \over \fermi\pars{1}}^{1/n}
\sim
\bracks{1 + {\fermi'\pars{1} \over \fermi\pars{1}n}}^{1/n}
=
\exp\pars{{1 \over n}\,\ln\pars{1 + {\fermi'\pars{1} \over \fermi\pars{1}n}}}
\sim
\exp\pars{{1 \over n}\,\bracks{{\fermi'\pars{1} \over \fermi\pars{1}n}}}
$$
So,
$$\color{#0000ff}{\large%
\lim_{n \to \infty}\bracks{\fermi\pars{1 + 1/n} \over \fermi\pars{1}}^{1/n}
=1}
$$
