I’ve been kicking this problem around for a couple of days, just as an exercise, and think I have a solution that does not depend on derivatives, but does depend on the fundamental definition for e:
e = lim (1+1/h)^h h->INF
and the associated knowledge of limits (e.g. as x->0, 1/x -> INF). Frankly if you’re dealing with a number like e (as the base for the natural logarithms), I think you you have to concede these basic terms.
1. (ln x)/x has a local maximum at x=e: Consider the function y = ln(x)/x. Choose two points on this curve: (x,ln(x)/x) and (ax,ln(ax)/(ax)) where a>1. Given x, what value of a will make the two y-values of these points the same (i.e. such that a line drawn between the two would be horizontal)?
Taking a>1 and x>0, the two y-values will be identical only if
ln(ax)/ax = ln(x)/x => ln(ax) = ln(x^a) => ax = x^a => a = x^(a-1) => x = a^[1/(a-1)]
Now make the variable change h = 1/(a-1) => a = 1+1/h:
x = (1+1/h)^h
Now let a -> 1 => h -> INF: By definition, x will approach e from below and ax will approach e from above. Since x and ax are approaching each other, the horizontal line connecting the two points becomes a horizontal tangent touching only one point, so x=e is an extremum; in fact it is a local maximum since the surrounding values ln(1)/1=0 and ln(e^2)/e^2=2/e^2 are both less than ln(e)/e.
**2. (ln x)/x is bounded from above: **It is relatively simple to show that ln(x+1) <= x by appealing only to the definition of e and limits: Take the function ln (x+1), draw a line connecting (x, ln(x+1)) and (x+a, ln(x+a+1)) where x>0 and a>0. The slope of this line must be (1/a)*ln((x+a+1)/(x+1)), which for fixed a>0 decreases with increasing x. This indicates that for two finite line segments of this type with fixed x-axis width a, the one closer to the origin always has a larger slope. If these two segments share a common endpoint (where one ends the other begins), this means they always meet at an angle pointing “upwards” on the graph, hence the function ln(x+1) is concave downward at every point.
Draw the specific line between (0,0) and (x, ln(x+1)). The slope of this line must be ln((1+a)^(1/a)). Making the variable change h=1/a gives ln((1+1/h)^h), and as above the argument of the logarithm tends toward e as h->INF (i.e. as a->0). Thus, this line becomes a tangent at (0,0) when its slope reaches 1. Because the curve is everywhere concave downward, the tangent cannot cross back over ln(x+1), so every y-value on the line y=x is equal to or greater than the corresponding value of y=ln(x+1) => x>=ln(x+1) (equal only when x=0). This of course implies that ln(x+1)/x is bounded from above for all x>-1 (excepting for the moment x=0), and a fortiori (ln x)/x is bounded for x>0.
Thus, we have proven that ln x/x has a local maximum at x=e, and that it is bounded from above for all values. Therefore, x=e produces the maximum value. QED, without derivatives. Perhaps a pointless exercise, but there it is…