Limit

1/x - 1/(e^x - 1)
= [(e^x - 1) - x]/[x(e^x - 1)]
= (e^x - 1 - x)/[x (e^x - 1)]
= (1 + x + x^2/2! + x^3/3! + ..... - 1 - x)/[x(e^x - 1)]
= (x^2/2! + x^3/3! + .....)/[x(e^x - 1]
= (x/2! + x^2/3! + x^3/4! + ......)/(x + x^2/2! + x^3/3! + .....)
= [x(1/2! + x/3! + x^2/4! + .....)/[x(1 + x/2! + x^2/3! + ....]
= (1/2! + x/3! + x^2/4! + .....)/(1 + x/2! + x^2/3! + ....)
So when x tends to 0, the expression tends to (1/2!)/(1) = 1/2.

infinite series is a gd try, but 因為有1/(e^x - 1)，所以諗緊可唔可以用例如lim x->0 (e^x-1)/x = 1 ge 公式

How about using infinite series?

e^x = 1 + x + x²/2 + x³/3! + ...

Your answer should be 1/2.