Answer
$$\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{e^x-1}\Big)=\frac{1}{2}$$
Work Step by Step
$$A=\lim_{x\to0^+}\Big(\frac{1}{x}-\frac{1}{e^x-1}\Big)$$
We unfortunately cannot do the replacement now since it would lead to the form $\infty-\infty$, which is a dead-end.
$$A=\lim_{x\to0^+}\frac{e^x-1-x}{x(e^x-1)}$$
$\lim_{x\to0^+}(e^x-1-x)=(e^0-1-0)=1-1=0$ and $\lim_{x\to0^+}[x(e^x-1)]=0(e^0-1)=0.$
We have an indeterminate form of $\frac{0}{0}$, which can be dealt with by L'Hospital's Rule:
$$A=\lim_{x\to0^+}\frac{(e^x-1-x)'}{[x(e^x-1)']}$$
$$A=\lim_{x\to0^+}\frac{e^x-1}{(e^x-1)+x(e^x-0)}$$
$$A=\lim_{x\to0^+}\frac{e^x-1}{e^x-1+xe^x}.$$
$\lim_{x\to0^+}(e^x-1)=e^0-1=1-1=0$ and $\lim_{x\to0^+}(e^x-1+xe^x)=e^0-1+0\times e^0=1-1+0=0.$
Another indeterminate form of $\frac{0}{0}$. We continue to take advantage of L'Hospital's Rule:
$$A=\lim_{x\to0^+}\frac{(e^x-1)'}{(e^x-1+xe^x)'}$$
$$A=\lim_{x\to0^+}\frac{e^x}{e^x-0+(e^x+xe^x)}$$
$$A=\lim_{x\to0^+}\frac{e^x}{2e^x+xe^x}$$
$$A=\frac{e^0}{2e^0+0\times e^0}$$
$$A=\frac{1}{2\times1+0}$$
$$A=\frac{1}{2}$$
