The given expression is the following limit:

$\lim_{{x \to a}} \frac{{(x - a)^{42} - 7x + 7a}}{{x - a}}.$

Step-by-step solution:

1. Direct Substitution: Substituting $x = a$ directly into the expression results in an indeterminate form $\frac{0}{0}$, so we need to apply L'Hôpital's Rule or simplify the expression in another way.

2. Simplification of the Numerator: The numerator consists of two terms: $(x - a)^{42}$ and $-7x + 7a$.

   • $(x - a)^{42}$ vanishes when $x = a$.
   • $-7x + 7a = -7(x - a)$, so the expression simplifies as follows:

   $\frac{{(x - a)^{42} - 7x + 7a}}{{x - a}} = \frac{{(x - a)^{42} - 7(x - a)}}{{x - a}}.$

3. Factoring the Numerator: Factor $(x - a)$ from the numerator:

   $\frac{{(x - a) \left( (x - a)^{41} - 7 \right)}}{{x - a}}.$

   Cancel the common factor $(x - a)$ (since $x \neq a$):

   $(x - a)^{41} - 7.$

4. Evaluating the Limit: Now, substitute $x = a$ into the simplified expression:

   $\lim_{{x \to a}} \left( (x - a)^{41} - 7 \right) = 0^{41} - 7 = -7.$

Thus, the value of the limit is $\boxed{-7}$.

Would you like more details on any specific part of this solution? Here are some related questions:

1. How does L'Hôpital's Rule work for indeterminate forms like this?
2. Can this limit be evaluated by expanding the binomial terms instead?
3. What happens if the power in the original expression was lower (e.g., $(x - a)^2$)?
4. Why is canceling $(x - a)$ allowed even though $x = a$ gives 0?
5. Could we apply a Taylor expansion to simplify this limit?

Tip: For limits involving indeterminate forms like $\frac{0}{0}$, simplifying the expression (factoring or applying L'Hôpital's Rule) is often the first step.