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To solve the given limit problem:

\[
\lim_{{x \to 5}} \frac{x^2 - 25}{x + 5}
\]

### Step 1: Factor the numerator
The numerator \(x^2 - 25\) is a difference of squares and can be factored as:

\[
x^2 - 25 = (x - 5)(x + 5)
\]

So, the expression becomes:

\[
\lim_{{x \to 5}} \frac{(x - 5)(x + 5)}{x + 5}
\]

### Step 2: Simplify the expression
For \(x \neq -5\), the term \((x + 5)\) cancels out from the numerator and denominator:

\[
\lim_{{x \to 5}} (x - 5)
\]

### Step 3: Evaluate the limit
Now, substitute \(x = 5\) into the simplified expression:

\[
5 - 5 = 0
\]

Thus, the limit is:

\[
\boxed{0}
\]

Would you like more details on any of the steps?

Here are 5 related questions for further exploration:
1. What is the limit of \(\lim_{{x \to -5}} \frac{x^2 - 25}{x + 5}\)?
2. How can the factorization of polynomials help in solving limits?
3. What happens if you substitute \(x = 5\) directly in the original expression without simplifying first?
4. How do you handle limits that involve indeterminate forms like \(\frac{0}{0}\)?
5. What is the difference between factoring and using L'Hopital's Rule for solving limits?

**Tip:** Always check for the possibility of factoring expressions when solving limits, especially when encountering indeterminate forms.

Calcule el límite de la siguiente función: lim (x→5) (x^2 - 25) / (x + 5)

Learn how to solve the limit lim (x→5) (x^2 - 25) / (x + 5) using algebraic factoring and cancellation techniques. This detailed step-by-step solution is ideal for high school students learning limits and indeterminate forms. Suitable for Grades 10-12.

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