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I'll analyze it now.From the image, it looks like you have the following problem:

$\text{Evaluate the limit:} \quad \lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

This is a standard limit problem. Let's solve it.

Step 1: Factor the numerator

The expression in the numerator is a difference of squares:

$x^2 - 25 = (x - 5)(x + 5)$

Step 2: Simplify the expression

Now, substitute the factorized form into the original expression:

$\frac{(x - 5)(x + 5)}{x - 5}$

Step 3: Cancel out the common factor

Since $x - 5$ appears in both the numerator and the denominator, we can cancel it out (as long as $x \neq 5$):

$\frac{(x - 5)(x + 5)}{x - 5} = x + 5$

Step 4: Evaluate the limit

Now that the expression is simplified, we can directly substitute $x = 5$ into the simplified expression:

$x + 5 = 5 + 5 = 10$

Thus, the value of the limit is:

$\boxed{10}$

Feel free to ask for more details or if you have any questions!

Here are five related questions to explore:

1. How do you solve limits involving indeterminate forms like $0/0$?
2. Can you always cancel terms like $(x - 5)$ when simplifying limits?
3. What is L'Hopital's Rule, and when would you use it?
4. What are the conditions for applying the difference of squares method in limits?
5. How do you evaluate limits at infinity?

Tip: Always check if you can factor expressions in limit problems. Factoring often helps simplify and remove indeterminate forms.