2
$\begingroup$

I need to prove that $f(x) = \frac{\ln{x}}{x}$ has a maximum in $(0,\infty)$.

Is this a valid solution:

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

  2. 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 $

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

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

  5. $M= \max(M_1, f(N_1))$ is the wanted maximum

Is this proof correct? Is there a better way?

  • 0
    If you can differentiate, you can just look at the derivative. Also, in your proof, it might be useful to note that $\frac{\ln e}{e} = 1/e > 0$. – 2017-01-13
  • 0
    I still can not use that. Should it be noted in the explanation why the maximum can't be in the $(0,1)$ interval? – 2017-01-13
  • 0
    Your proof is correct, and I think that it is the standard proof one could use when there are no original ideas. Anyway, good job. – 2017-01-13
  • 0
    do you just want to prove there is a maximum? That follows just from continuity and the fact that the limit at $\infty$ is zero. – 2017-01-13
  • 0
    @BeniBogosel how so? Is there a known statement that makes it trivial? – 2017-01-13
  • 1
    Before $x=1$ the function is negative so we know that the maximum is in $[1,\infty)$. Since the function has limit zero at $\infty$ and is continuous it will have a maximal value. The steps you wrote are proof of exactly that. You know that for $x$ large enough the function is smaller than $f(e)$, for example, so you can look for the max in a bounded closed interval. In that interval you apply the result concerning continuous functions on compact sets which touch their extreme values... – 2017-01-13

3 Answers 3

1

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 $(1,\infty)$.

  • 0
    How can you say for sure without proving that the maximum is somewhere in $[1,N]$, that's my problematic part. If I could say that, the Extreme value theorem would solve this right away – 2017-01-13
  • 1
    @S.Peter The function *restricted* to $[1,N]$ has a maximum. Added. – 2017-01-13
  • 1
    @S.Peter Because any value of the function beyond $x=N$ is inferior to a particular value that the function takes on $[1,N]$, namely $1/e$ (as $N$ has been chosen this way), hence the maximum needs to be in $[1,N]$. – 2017-01-13
  • 0
    It makes perfect sense, just taking a specific case here with $e$ makes it much clearer "There exists $N>e$ such that, for $x>N$, $f(x)<1/e$." I just want to make sure I understand that one. It is said due to the fact that $lim_{x\to\infty}f(x)=0$?, correct? – 2017-01-13
  • 1
    @S.Peter Yes; the definition says there is some $N_0>0$ such that $|f(x)|<1/e$, for $x>N_0$; then take $N=\max\{e+1,N_0\}$, if you want to be fussy. – 2017-01-13
4

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 $x=e$.

  • 0
    Few questions: how do you know that there is an equality only if $x=e$ and moreover, I still can't see how this answers my question, hopefully you can explain it into more details – 2017-01-13
  • 2
    @S. Peter $x^{1/x}$ is nothing but the exponential of $f(x)$. Since exponential is monotonic, $f$ has a maximum at $x=e$ if and only if $x^{1/x}$ has. – 2017-01-13
1

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.