First a couple of remarks:
I will asume that #log x# is #log_e x#
Clearly the expression does not have a single 'value' for #x in (0,oo)# I will analyse the function #f(x) = lnx/x#
First consider the domain of #f(x)#
Since #lnx# is defined #forall x in RR >0# The domain of #f(x)# is #(0, +oo)# Which is the interval we have been asked to consider.
Now, let's consider the values of #f(x)# at it's limits
#lim_(x->0+) lnx/x = lim_(x->0+ )lnx * lim_(x->0+) 1/x#
#lim_(x->0+ )lnx =-oo#
# lim_(x->0+) 1/x =+oo#
#:.lim_(x->0+) lnx/x =-oo#
and
#lim_(x->oo) lnx/x -> oo/oo#
#lim_(x->oo) lnx/x = lim_(x->oo) (d/dx lnx)/(d/dx x)# [L'Hôpital's rule]
#= lim_(x->oo) (1/x)/1 = lim_(x->oo) 1/x =0#
So, we have now determined the values of #f(x)# at its limits to be #(0, -oo)# and #(+oo, 0)#
Next, we will find the critical points of #f(x)#
#f'(x) = (x*(1/x)-lnx*1)/x^2# [Quotient rule]
#= (1-lnx)/x^2#
For critical points #f'(x)=0#
#(1-lnx)/x^2=0#
#lnx =1#
#x = e^1 =e approx 2.7183#
#f(e) = lne/e = 1/e approx 0.3679#
To determine the nature of #f(e)#
#f''(x) = (x^2(1/x)-(1-lnx)*2x)/x^4# [Quotient rule]
#= (-x-(1-lnx)*2x)/x^4#
#:. f''(e) = (-e-(1-1)*2e)/e^4 = (-e)/e^4 =-1/e^3 <0#
Since #f''(e)<0 -> f(e) = f_"max"# (i.e. a maximum value of #f(x)#)
Since #f(x)# has a maximum value #>0# it must have a finite zero.
#f(x) =0 -> lnx/x=0#
#:. lnx =0#
#x=e^0 =1#
So, we now have determined that #f(x)# has a maximum at #(e,1/e)# and a finite zero at #(1,0)#
Combining this information with the values of the limits of #f(x)# above we are able to graph #f(x)# as below.
graph{lnx/x [-1.11, 11.38, -4.24, 2]}
[N.B. In practice we would probably need to plot a few extra points]
