Answer
$$\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x-\sin x}=2$$
Work Step by Step
$$A=\lim_{x\to0}\frac{e^x-e^{-x}-2x}{x-\sin x}$$
$\lim_{x\to0}(e^x-e^{-x}-2x)=e^0-e^{-0}-2\times0=1-1-0=0$ and $\lim_{x\to0}(x-\sin x)=0-\sin0=0,$
which is the indeterminate form $\frac{0}{0}$. L'Hospital's Rule can be used:
$$A=\lim_{x\to0}\frac{(e^x-e^{-x}-2x)'}{(x-\sin x)'}$$
$$A=\lim_{x\to0}\frac{e^x-e^{-x}(-x)'-2}{1-\cos x}$$
$$A=\lim_{x\to0}\frac{e^x-e^{-x}(-1)-2}{1-\cos x}$$
$$A=\lim_{x\to0}\frac{e^x+e^{-x}-2}{1-\cos x}$$
$\lim_{x\to0}(e^x+e^{-x}-2)=e^0+e^{-0}-2=1+1-2=0$ and $\lim_{x\to0}(1-\cos x)=1-\cos0=1-1=0.$
Another indeterminate form: $\frac{0}{0}$. We use L'Hospital's Rule one more time:
$$A=\lim_{x\to0}\frac{(e^x+e^{-x}-2)'}{(1-\cos x)'}$$
$$A=\lim_{x\to0}\frac{e^x+e^{-x}(-x)'}{0-(-\sin x)}$$
$$A=\lim_{x\to0}\frac{e^x-e^{-x}}{\sin x}$$
$\lim_{x\to0}(e^x-e^{-x})=e^0-e^{-0}=1-1=0$ and $\lim_{x\to0}\sin x=\sin0=0.$
Apparently, this indeterminate form of $\frac{0}{0}$ still has one last shot. We still can use L'Hospital's Rule:
$$A=\lim_{x\to0}\frac{(e^x-e^{-x})'}{(\sin x)'}$$
$$A=\lim_{x\to0}\frac{e^x+e^{-x}}{\cos x}\hspace{1cm}(\text{This is finally over.})$$
$$A=\frac{e^0+e^{-0}}{\cos 0}$$
$$A=\frac{1+1}{1}$$
$$A=2$$
