Let's solve the given limit problem step by step.

We are tasked with finding the limit:

$\lim_{{x \to 8^-}} \frac{{x - 8}}{{x^2 - 64}}$

Step 1: Simplify the expression.

We start by recognizing that the denominator $x^2 - 64$ is a difference of squares: $x^2 - 64 = (x - 8)(x + 8)$

So, the expression becomes: $\frac{{x - 8}}{{(x - 8)(x + 8)}}$

Step 2: Cancel common factors.

For $x \neq 8$, we can cancel the common factor of $x - 8$ in the numerator and denominator: $\frac{{1}}{{x + 8}}$

Step 3: Evaluate the limit.

Now, we can directly substitute $x = 8$ into the simplified expression: $\lim_{{x \to 8^-}} \frac{1}{x + 8} = \frac{1}{8 + 8} = \frac{1}{16}$

Final Answer:

The limit is $\frac{1}{16}$.

Do you need further details or explanations?

Here are 5 related questions for practice:

1. How would you compute the limit as $x \to 8^+$?
2. What happens if you try to calculate the limit as $x \to 8$ without simplifying the expression?
3. How does the concept of factoring help in simplifying limits with indeterminate forms?
4. Can you explain why direct substitution initially leads to an indeterminate form in this case?
5. How does this example illustrate the difference between one-sided and two-sided limits?

Tip: Always check for factorable expressions in limits that result in indeterminate forms like $\frac{0}{0}$. Simplification often resolves the indeterminacy.