Prove that for every $x>0$, it follows that $\frac{\ln x}{x} \leq\frac{1}{e}$ I need to prove that for every $x&gt;0$ $$\frac{\ln x}{x} \leq\frac{1}{e}$$

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