Evaluate the following limit: $\lim _{x \rightarrow \infty}\left(\frac{x-1}{x}\right)^{x}$ $\lim _{x \rightarrow \infty}\left(\frac{x-1}{x}\right)^{x}=e$ $\lim _{x \rightarrow \infty}\left(\frac{x-1}{x}\right)^{x}=-1$ $\lim _{x \rightarrow \infty}\left(\frac{x-1}{x}\right)^{x}=\frac{1}{e}$ $\lim _{x \rightarrow \infty}\left(\frac{x-1}{x}\right)^{x}=1$

Step 1 ：Evaluate the following limit: \(\lim _{x \rightarrow \infty}\left(\frac{x-1}{x}\right)^{x}\)

Step 2 ：The limit is in the form of \([1 + \frac{f(x)}{x}]^x\) which is a standard limit and its value is \(e^{f(x)}\). Here, \(f(x) = -1\). So, the limit should be \(e^{-1}\) or \(\frac{1}{e}\).

Step 3 ：Final Answer: The value of the limit is \(\boxed{\frac{1}{e}}\) or approximately \(\boxed{0.36787944117144233}\).