How do you find the Limit of lnxlnx as x approaches 0?

1 Answer
Feb 6, 2017

lim_(x->0) lnx = -oo

Explanation:

First we prove that ln(x) is monotone increasing.

Consider:

x_1, x_2 in RR^+ with x_2 > x_1

ln x_2 = ln (x_2/x_1*x_1) = ln( x_2/x_1) +ln x_1

x_2 > x_1 => x_2/x_1 > 1 => ln(x_2/x_1) > 0 => ln(x_2) > ln(x_1)

which proves the point.

Since it is monotone increasing lnx has a limit for x->oo and since the function is not bounded this limit must be +oo, so:

lim_(x->oo) lnx = +oo

Now note that:

ln(1/x )= -ln x

and that as the logarithm is defined only for x > 0

lim_(x->0) lnx = lim_(x->0^+) lnx

Substitute now y=1/x

lim_(x->0^+) lnx = lim_(y->oo) ln(1/y) = lim_(y->oo) -ln(y) = -oo