lim x + 1 / x kuadrat - 4x -5 =. lim = x adalah -1

We are tasked with finding the following limit:

\[
\lim_{x \to -1} \frac{x + 1}{x^2 - 4x - 5}
\]

### Step 1: Substitute \( x = -1 \) into the expression.

First, we substitute \( x = -1 \) into the numerator and denominator:

#### Numerator:
\[
x + 1 = (-1) + 1 = 0
\]

#### Denominator:
\[
x^2 - 4x - 5 = (-1)^2 - 4(-1) - 5 = 1 + 4 - 5 = 0
\]

So, directly substituting gives us a form \( \frac{0}{0} \), which is indeterminate. This indicates that we need to factor and simplify the expression.

### Step 2: Factor the denominator

The denominator is a quadratic expression. Let's factor it:

\[
x^2 - 4x - 5 = (x - 5)(x + 1)
\]

### Step 3: Simplify the expression

Now, rewrite the limit with the factored denominator:

\[
\lim_{x \to -1} \frac{x + 1}{(x - 5)(x + 1)}
\]

Notice that \( x + 1 \) appears in both the numerator and denominator, so we can cancel it:

\[
\lim_{x \to -1} \frac{1}{x - 5}
\]

### Step 4: Substitute \( x = -1 \)

Now, substitute \( x = -1 \) into the simplified expression:

\[
\frac{1}{-1 - 5} = \frac{1}{-6}
\]

### Final Answer:
\[
\lim_{x \to -1} \frac{x + 1}{x^2 - 4x - 5} = -\frac{1}{6}
\]

Would you like more details or have any further questions?

---

Here are 5 related questions for further practice:

1. What is the limit \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \)?
2. How would you solve \( \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \)?
3. Can you factor and simplify \( x^2 - 5x + 6 \) and find the limit as \( x \to 2 \)?
4. What is the behavior of \( \lim_{x \to -2} \frac{x^2 + 2x}{x + 2} \)?
5. How do we handle limits that result in indeterminate forms like \( \frac{0}{0} \)?

**Tip:** Always factor expressions when you encounter the indeterminate form \( \frac{0}{0} \), as it often simplifies the limit.

Find the solution for the limit problem \( \lim_{x \to -1} \frac{x + 1}{x^2 - 4x - 5} \). This detailed explanation covers the process of factorization and simplification of indeterminate forms, making it perfect for high school calculus students (Grades 11-12).

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