How do you find the limit of #[1/ln(x)] - [1/x-1]# as x approaches 1?

1 Answer
May 24, 2016

The limit does not exist.

Explanation:

As #xrarr1#, the second term goes to #0#.

#lim_(xrarr1)(1/x-1) = 1/1-1=0#

So the limit depends entirely on the limit of the first term.

As #xrarr1#, the natural logarithm goes to #0#,

#lim_(xrarr1)lnx = 0#

Therefore, #1/lnx# is either increasing or decreasing without bound (going to #oo# or to #-oo#) depending on whether #x# is to the right or left of #1#.

For #0 < x < 1#, we have #lnx < 0# so, #lim_(xrarr1^-)1/lnx = -oo#.

On the other hand, for #1 < 1#, we have #lnx > 0#, so #lim_(xrarr1^+)1/lnx = oo#.

The two-sided limit does not exist.