The problem presents the following limit:

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

To solve for this limit, let's analyze the behavior of the function as $x$ approaches 8.

1. Direct Substitution: If we try to substitute $x = 8$ directly into the expression, we get:

$\frac{8^2}{(8 - 8)^2} = \frac{64}{0}$

This leads to division by zero, which means we need to analyze the behavior as $x$ approaches 8 from both sides.

2. Behavior as $x \to 8^+$ (approaching from the right): When $x$ is just slightly greater than 8, $(x - 8)^2$ becomes a small positive number, making the denominator very small and positive. Since $x^2 = 64$, the fraction will tend towards positive infinity.

3. Behavior as $x \to 8^-$ (approaching from the left): Similarly, when $x$ is slightly less than 8, $(x - 8)^2$ is still a small positive number, making the denominator again small and positive. The fraction will still tend towards positive infinity.

Since the denominator approaches zero positively from both sides and the numerator stays positive, we conclude:

$\lim_{x \to 8} \frac{x^2}{(x - 8)^2} = +\infty$

Therefore, the correct answer is $\infty$.

Would you like further explanation or details on this topic?

Here are 5 related questions you can explore:

1. How would the limit change if the numerator were $x^3$ instead of $x^2$?
2. What is the difference between this limit and a limit involving division by a linear term like $(x - 8)$?
3. How do you evaluate limits involving higher powers in the denominator or numerator?
4. What are indeterminate forms, and how do they relate to limits?
5. How would you apply L'Hopital's Rule to solve similar limits?

Tip: Always examine the sign of the denominator as you approach a point of discontinuity to predict the behavior of the limit effectively!