The given expression is a limit: $\underset{x\to 0}{\lim }\frac{\sin x}{e^x-1}$. To evaluate it, we recognize it as a 0/0 indeterminate form. We can apply L'Hôpital's Rule, which states that if the limit of f(x)/g(x) is indeterminate, it equals the limit of f'(x)/g'(x).

The derivative of the numerator, sin x, is cos x. The derivative of the denominator, e^x - 1, is e^x. Therefore, the limit becomes: $\underset{x\to 0}{\lim }\frac{\cos x}{e^x}=\frac{\cos 0}{e^0}=\frac{1}{1}=1$.

Final Answer: $1$