Verified answer

Let's look at it as

lim x->1 of { [x - 1 - ln(x)] / [ln(x)*(x-1)] }

The derivative of the numerator is 1 - 1/x.

The derivative of the denominator is

ln(x) + (x-1)/x = ln(x) + 1 - 1/x.

Both of these still go to zero when x goes to 1,

so we will apply L'Hopital's Rule a second time,

trying to find

lim x->1 { (1 - 1/x) / [ln(x) + 1 - 1/x] }.

The derivative of the numerator is 1/x^2.

The derivative of the denominator is 1/x + 1/x^2.

So the limit should be 1/2.

To check this, I'll let x be 1.00001. Then

(1/ln(1.00001)) - 1/(0.00001)

= 0.49999851 by the Google calculator,

so my answer of 1/2 looks good!

Lt (x→1) [1/lnx – 1/(x-1)]

Using L-Hospital rule

Lt (x→1) [0/1/x – (-1/(x-1)²]

Lt (x→1) [1/(x-1)²