Evaluating the limit with power series

lim lnx/(x-1)
x->1

I only got to ln(x)(-1/1-x) using the geometric series 1/(1-x) = Σ (n=0->infinity) x^n:
-ln(x)Σ (n=0->infinity) x^n

Can someone help me the steps after? Thanks

Starting with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n:

Integrate both sides from 0 to t:
- ln(1 - t) = Σ(n = 0 to ∞) t^(n+1)/(n+1)
==> ln(1 - t) = -Σ(n = 0 to ∞) t^(n+1)/(n+1).

Let t = 1 - x:
ln x = -Σ(n = 0 to ∞) (1 - x)^(n+1)/(n+1).
......= (x - 1) - (x - 1)^2/2 + ...

Hence,
lim(x→1) ln(x)/(x - 1)
= lim(x→1) [(x - 1) - (x - 1)^2/2 + ...]/(x - 1)
= lim(x→1) [1 - (x - 1)/2 + ...]
= 1 - 0
= 1.

I hope this helps!