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

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Answer 1 · 2017-02-06T01:36:09

#lim_(x->0) lnx = -oo#

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#