Answer
$\dfrac{1}{2}$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0} (\dfrac{1}{x}-\dfrac{1}{e^x-1})=\lim\limits_{x \to 0} \dfrac{e^x-x-1}{x(e^x-1)}$
We can see that the numerator and denominator have a limit of $0$, so the limit shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0} \dfrac{e^x-x-1}{x(e^x-1)}=\lim\limits_{x \to 0} \dfrac{e^x-1}{xe^x+e^x-1}\\= \lim\limits_{x \to 0} \dfrac{1}{x+2}=\dfrac{1}{2}$
