We have to prove that for every #x_0 in (0,+oo)# we have:

#lim_(x->x_0) log x = log x_0#

that is for every #epsilon > 0# we can find #delta_epsilon > 0# such that for every #x in (x_0-delta_epsilon, x_0+delta_epsilon)# we have:

#abs (log x -log x_0) < epsilon#.

Now we can restrict ourselves to natural logarithms since:

#log_a x = ln x /ln a#

and a continuous function multiplied by a constant is still continuous.

Then we note the following properties of logarithms:

#(i) ln(x)-ln(x_0) = ln (x/x_0)#

#(ii) x_1 < x_2 => ln x_1 < ln x_2, " since "e>1#

Now, given #x_0 in (0,+oo)# and #epsilon >0# we choose:

#delta_epsilon = min(x_0(e^epsilon -1), x_0(-e^-epsilon+1)) #

Since #epsilon > 0#, #e^epsilon > 1# and #e^(-epsilon) < 1# so both number are positive.

Now we consider #x in (x_0-delta_epsilon, x_0+delta_epsilon)# and we distinguish the two cases:

(1) #x> x_0#

#abs (ln x -ln x_0) = ln x - ln x_0 = ln (x/x_0) < ln ((x_0 + x_0(e^epsilon -1))/x_0)=ln e^epsilon = epsilon#

(2) #x< x_0#

#abs (ln x -ln x_0) = ln x_0 - ln x = ln (x_0/x) < ln (x_0 / (x_0- x_0(-e^-epsilon+1)))=ln(1/e^-epsilon) = ln (e^epsilon) = epsilon#

In either case we have:

#abs (ln x -ln x_0) < epsilon#

and the continuity is proved.