prove that $f(x) = \frac{\ln{x}}{x}$ has a maximum in $(0,\infty)$ without derivatives. I need to prove that $f(x) = \frac{\ln{x}}{x}$ has a maximum in $(0,\infty)$.
Is this a valid solution:


*

*Explaining that if $f(x)$ has a maximum it can't be in $(0,1)$ since $f(x)<0$ in that section

*I showed that $\lim_{x\to\infty}\frac{\ln{x}}{x} = 0,$ therefore by definition of limit at infinity, for every $\epsilon$ exists $N>0$ which for every $x>N$ this holds: $|f(x) - 0| < \epsilon $

*Let $\epsilon_1 \in R$. therefore, exists $N_1$ which for every $x>N_1, f(x)< \epsilon$

*Looking at $[1,N_1]$, using the Extreme value theorem, there is a maximum in that bounded interval, say $M_1$.

*$M= \max(M_1, f(N_1))$ is the wanted maximum
Is this proof correct? Is there a better way?
 A: Here is a proof by magic,
not original with me,
if we know that
$e^x \ge 1+x$
with equality if and only if
$x = 0$.
$e^{\frac{x}{e}-1}
=e^{\frac{x-e}{e}}
\ge 1+\frac{x-e}{e}
=\frac{x}{e}
$
with equality if and only if
$x=e$.
Therefore,
if $x \ne e$,
$e^{\frac{x}{e}}
> x$
so that
$e^x > x^e$
or
$e^{1/e}
> x^{1/x}
$.
Therefore the unique maximum
is at $x=e$.
A: Minor note: $N$ is in $[1,N]$ so $\max(M_1,f(N))=M_1$.   
The real problem is that you haven't shown $M_1$ is the maximum.  Based on what you wrote it need not be, e.g. if $N$ was chosen based on $\varepsilon=80$.  It would suffice to choose $N$ based on $\varepsilon=f(5)$, say.  Then you would know that $M\geq f(5)> f(x)$ for all $x>N$.  
If you have the means to prove the function is decreasing for sufficiently large $x$, it would suffice to choose $N$ such that $f$ is decreasing on $[N,\infty)$, with no need to consider the limit.
A: 
Is this proof correct?

Almost. Point 3 is rather obscure.
For $x\in(0,1)$, we have $f(x)<0$. Good: no maximum there, we can concentrate on $[1,\infty)$ where $f(x)\ge0$.
We have $f(e)=1/e>0$. There exists $N>e$ such that, for $x>N$, $f(x)<1/e$.
The function $f$, restricted to $[1,N]$, has a maximum on $[1,N]$, say at $x_0$. Since $e\in [1,N]$, we have $f(x_0)\ge f(e)=1/e$. Therefore $f(x_0)$ is the required global maximum, since $f(x_0)\ge f(x)$ for $x\in[1,N]$ and also for $x>N$.
Note. Instead of $e$, which is where the function actually attains its maximum, you could choose any point in $(1,\infty)$.
