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I need to prove that for every $x>0$ $$\frac{\ln x}{x} \leq\frac{1}{e}$$

I tried to show that when the limit goes to $+\infty$ the function $\frac{\ln x}{x} -\frac{1}{e}$ goes to $-1/e$ but that of course doesn't guarantee it won't suddenly “jump” far and beyond somewhere along the way and then come back.

When I take the derivative I get that for some of the range the function is going up rather than down, so that it being smaller than $0$ cannot be guaranteed. As a result don't I know how to prove this.

egreg
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Bak1139
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  • Mathjax/Latex tip: Instead of "<=". Type "\leq" (less or equal) with dollar signs around it. It will look like this: $\leq$ –  Jan 05 '18 at 13:30
  • you just need to show that $x^e\le e^x$ –  Jan 05 '18 at 13:37
  • you can have a look: https://math.stackexchange.com/questions/1619911/why-ex-is-always-greater-than-xe/1619966#1619966 – Domates Jan 05 '18 at 18:28

6 Answers6

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defining $$f(x)=\frac{x}{e}-\ln(x)$$ then we get: $f(x)$ has a Minimum in $x=e$ and $$\lim_{x\to 0^+}f(x)=+\infty$$ and $$\lim_{x\to \infty}f(x)=+\infty$$ thus we have $$\frac{\ln(x)}{x}\le \frac{1}{e}$$ for $x>0$

Dr. Sonnhard Graubner
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Let $f(x)=\ln x -\frac{x}e$. What you wish to prove is equivalent to $f(x)\leq 0$ for all $x<0$.

Notice that $f$ is continuous and $\lim_{x\to0^+}f(x)= \lim_{x\to+\infty}f(x)=-\infty$, so $f$ must attain a maximum on $(0,+\infty)$. (Why? Can you show this?)

Since $f$ is differentiable, the maximum must be a solution to $f'(x)=0$, that is, a solution to $\frac1x-$$\frac1e=0$. It's easy to see that the only solution is $x=e$. It follows that $f(e)=0$ is the maximum, which concludes the proof. $\square$

Fimpellizzeri
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For $x=e$ we have obtain equality. If we take $x\neq e$, for $x>e$(similarly for $x<e$, it can be shown) by MVT there exists $x_0 \in (x,e)$ such that \begin{align*} \frac{\ln e-\ln x}{e-x}&=\frac{1}{x_0}\\ \end{align*}

Hence we have \begin{align*} \frac{1-\ln x}{e-x}&=\frac{1}{x_0}\\ & > \frac{1}{e}\\ \end{align*} and

$$1-\frac{e-x}{e}>\ln x$$

namely $$\frac{1}{e}>\frac{\ln x}{x}.$$

Domates
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Consider $f(x)=\frac{\ln x}{x}$ with $$ f'(x)=\frac{1-\ln x}{x^2} $$ so that $f$ has a maximum at $x=e$; since $f(e)=1/e$, we get the claim.

egreg
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Let $f(x) = \frac {ln (x)} { x}.$

$f'(x) = \frac{1}{x^2}-\frac {ln (x)}{x^2}$ $=\frac {1}{x^2} (1-ln (x)) $

For $0 <x <e $, $f'(x) >0$. This means that $f (e)> f (x)$

For $x>e $, $f'(x)<0$. This means that $f (x) < f(e)$

For $x=e $, both sidess of the given equation are equal, hence the equality sign.

Since $f (e) = \frac {1}{e}$,

For all $x>0$, $ \frac {ln(x) }{x } \leq \frac {1}{e} $

dezilim
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$$ \frac{ln(x)}{x} \leq \frac{1}{e} \iff x^e \le e^x. e^x \in C(\mathbb R), x^e \in C(\mathbb R). $$ $e^x \gt x^e, x \in [0,e)$ and $(e,+\infty)$ because both of them are continious and have only one point of intersaction (e), after that point $e^x \gt x^e$ as before, so $\frac{ln(x)}{x} \leq \frac{1}{e}$ is proofed.

Sharlotta Neimor
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  • What a joke ?? so if two continuous functions have one common point then such inequality holds? –  Jan 05 '18 at 14:02
  • not really, we have two functions, both of them are continuous. They have one point of intersection and after this point and before this point one is greater then another, so that such inequality holds. – Sharlotta Neimor Jan 05 '18 at 14:08