The given limit is:

$\lim_{b \to 6} \frac{\frac{1}{b} - \frac{1}{6}}{b - 6}$

This expression is an indeterminate form of type $\frac{0}{0}$, so we can apply L'Hopital's Rule. This rule states that if a limit results in an indeterminate form, we can differentiate the numerator and denominator separately and then take the limit.

Step-by-Step Solution:

1. Differentiate the numerator $\frac{1}{b} - \frac{1}{6}$ with respect to $b$:

   • The derivative of $\frac{1}{b}$ is $-\frac{1}{b^2}$.
   • The derivative of $\frac{1}{6}$ is $0$ because it is a constant.

   So, the derivative of the numerator is $-\frac{1}{b^2}$.

2. Differentiate the denominator $b - 6$ with respect to $b$:

   • The derivative of $b - 6$ is $1$.

   Therefore, the limit becomes:

   = \lim_{b \to 6} -\frac{1}{b^2}$$

3. Substitute $b = 6$ into the expression: $-\frac{1}{6^2} = -\frac{1}{36}$

Thus, the value of the limit is: $\boxed{-\frac{1}{36}}$

Would you like further clarification or additional details on this method?

Here are five related questions for you to explore:

1. How does L'Hopital's Rule work for indeterminate forms other than $\frac{0}{0}$?
2. What other methods can be used to evaluate limits without using L'Hopital's Rule?
3. How would you solve this limit using algebraic simplification rather than calculus?
4. Can L'Hopital's Rule be applied multiple times to solve a limit?
5. What is the significance of indeterminate forms in the study of limits?

Tip: Always check if a limit results in an indeterminate form before applying L'Hopital's Rule. Not all limits need this method for simplification!