Solution
= Solution
Combine the terms before taking their <limit>:
$$
\frac1x-\frac1{e^x-1}=\frac{e^x-1-x}{x(e^x-1)}.
$$
Using the <exponential function> <Maclaurin series>, the numerator is $x^2/2+O(x^3)$ and the denominator is $x^2+O(x^3)$. Consequently
$$
\boxed{\lim_{x\to0}\left(\frac1x-\frac1{e^x-1}\right)=\frac12}.
$$
The cancellation removes the two individual singularities. Subtracting their separate divergent limits would not determine the finite answer.