Find the following limit or state that it does not exist Assume \( b \) is a fixed real number. \( \lim _{x \rightarrow b} \frac{(x-b)^{60}-3 x+3 b}{x-b} \)

To find the limit \[ \lim_{x \rightarrow b} \frac{(x-b)^{60} - 3x + 3b}{x-b}, \] we start by substituting \( x = b \) into the expression. This gives us: \[ \frac{(b-b)^{60} - 3b + 3b}{b-b} = \frac{0 - 3b + 3b}{0} = \frac{0}{0}, \] which is an indeterminate form. To resolve this, we can apply L'Hôpital's Rule, which is applicable for limits of the form \( \frac{0}{0} \). First, we differentiate the numerator and the denominator separately. The denominator \( x - b \) differentiates to \( 1 \). Next, we differentiate the numerator: \[ f(x) = (x-b)^{60} - 3x + 3b. \] The derivative \( f'(x) \) is calculated as follows: 1. The derivative of \( (x-b)^{60} \) is \( 60(x-b)^{59} \). 2. The derivative of \( -3x \) is \( -3 \). 3. The derivative of \( 3b \) is \( 0 \). Thus, we have: \[ f'(x) = 60(x-b)^{59} - 3. \] Now we apply L'Hôpital's Rule: \[ \lim_{x \rightarrow b} \frac{(x-b)^{60} - 3x + 3b}{x-b} = \lim_{x \rightarrow b} \frac{60(x-b)^{59} - 3}{1}. \] Now we substitute \( x = b \): \[ \lim_{x \rightarrow b} (60(x-b)^{59} - 3) = 60(0) - 3 = -3. \] Thus, the limit is \[ \boxed{-3}. \]