To find the limit $\lim_{x \to 16} \frac{x - 16}{x^2 - 256}$, let's first try to directly substitute $x = 16$ and see what happens.

1. Direct Substitution:

$\frac{16 - 16}{16^2 - 256} = \frac{0}{256 - 256} = \frac{0}{0}$ This is an indeterminate form $\frac{0}{0}$, so we need to simplify the expression further before evaluating the limit.

2. Simplify the Expression:

Notice that the denominator $x^2 - 256$ can be factored as a difference of squares: $x^2 - 256 = (x - 16)(x + 16)$ So the limit expression becomes: $\lim_{x \to 16} \frac{x - 16}{(x - 16)(x + 16)}$ We can now cancel the common factor $(x - 16)$ in the numerator and denominator: $\lim_{x \to 16} \frac{1}{x + 16}$

3. Evaluate the Simplified Limit:

Now, substitute $x = 16$ into the simplified expression: $\frac{1}{16 + 16} = \frac{1}{32}$

So, the limit is: $\lim_{x \to 16} \frac{x - 16}{x^2 - 256} = \frac{1}{32}$

Would you like further details on any step or have any questions? Here are 5 related questions you might find useful:

1. How do you factor more complex polynomials to simplify limits?
2. What are some common indeterminate forms, and how are they resolved?
3. Can you always cancel terms directly when simplifying limits?
4. How would you use L'Hôpital's rule to solve this limit?
5. What happens if the limit expression involves higher powers of $x$?

Tip: Factoring is a powerful technique to simplify expressions, especially when dealing with limits that initially produce indeterminate forms.