Find the holes in the graph of the function \( f(x) = \frac{\sin{x} - 1}{x^2 - 1} \).

Step 1 ：Step 1: Find the values of \(x\) that cause the denominator to be zero, as these are the values that would cause the function to be undefined. Set \(x^2 - 1 = 0\), and solve for \(x\).

Step 2 ：\(x^2 - 1 = 0\)

Step 3 ：\(x^2 = 1\)

Step 4 ：\(x = \pm 1\)

Step 5 ：The function is undefined at \(x = 1\) and \(x = -1\).

Step 6 ：Step 2: Determine if these points are holes or vertical asymptotes by checking if the numerator also becomes zero at these points. If it does, then these are holes. Otherwise, they are vertical asymptotes.

Step 7 ：For \(x = 1\), \(\sin{1} - 1 = 0\) is not true. So, \(x = 1\) is not a hole.

Step 8 ：For \(x = -1\), \(\sin{-1} - 1 = 0\) is not true. So, \(x = -1\) is not a hole.

Step 9 ：So, there are no holes in the graph of the function.