Prove that for every $x>0$, it follows that $\frac{\ln x}{x} \leq\frac{1}{e}$

Question

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.

Answer 1

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. $\square$

Answer 2

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 $x>0$

Answer 3

$$ \frac{ln(x)}{x} \leq \frac{1}{e} \iff x^e \le e^x. e^x \in C(\mathbb R), x^e \in C(\mathbb R). $$ $e^x \gt x^e, x \in [0,e)$ 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.

Answer 4

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

Answer 5

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.

Answer 6

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} $