Compute the following limit using I'HI^opital's rule if appropriate. Use INF to denote $\infty$ and MINF to denote $-\infty$.
\[
\lim _{x \rightarrow \infty}\left(1-\frac{8}{x}\right)^{x}=\square
\]

Step 1 ：Let's denote the given expression as \(f = (1 - \frac{8}{x})^x\).

Step 2 ：We can rewrite the expression in a form that allows us to use L'Hopital's rule by taking the natural logarithm of the expression, which allows us to bring the exponent down. So, we have \(\ln(f) = \ln((1 - \frac{8}{x})^x)\).

Step 3 ：By applying L'Hopital's rule, we find that the limit of the natural logarithm of the expression as \(x\) approaches infinity is \(-\infty\), i.e., \(\lim_{x \rightarrow \infty} \ln(f) = -\infty\).

Step 4 ：Finally, we can find the original limit by taking the exponential of the result. Since the exponential of \(-\infty\) is 0, we have \(\lim_{x \rightarrow \infty} f = 0\).

Step 5 ：So, the limit of the given expression as \(x\) approaches infinity is \(\boxed{0}\).