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The expression you provided is:

\[
\lim_{{x \to -2}} \left( \frac{{x - 1}}{{x + 2}} - \frac{{x^2 + x + 1}}{{x^2 + 3x + 2}} \right)
\]

Let's solve it step by step.

### Step 1: Check for Direct Substitution

First, substitute \( x = -2 \) directly into the expression:

\[
\frac{{x - 1}}{{x + 2}} \quad \text{and} \quad \frac{{x^2 + x + 1}}{{x^2 + 3x + 2}}
\]

For the first term:

\[
\frac{{x - 1}}{{x + 2}} = \frac{{-2 - 1}}{{-2 + 2}} = \frac{{-3}}{0} \quad \text{(undefined due to division by zero)}
\]

For the second term:

\[
\frac{{x^2 + x + 1}}{{x^2 + 3x + 2}} = \frac{{(-2)^2 + (-2) + 1}}{{(-2)^2 + 3(-2) + 2}} = \frac{{4 - 2 + 1}}{{4 - 6 + 2}} = \frac{3}{0} \quad \text{(also undefined due to division by zero)}
\]

Since direct substitution leads to an indeterminate form \(\frac{0}{0}\), we will need to simplify the expression using algebraic techniques like factoring.

### Step 2: Simplify the Expression

The denominator \(x^2 + 3x + 2\) can be factored:

\[
x^2 + 3x + 2 = (x + 2)(x + 1)
\]

Now the original limit expression becomes:

\[
\lim_{{x \to -2}} \left( \frac{{x - 1}}{{x + 2}} - \frac{{x^2 + x + 1}}{{(x + 2)(x + 1)}} \right)
\]

### Step 3: Combine the Two Fractions

To combine the two terms into a single fraction, we'll need a common denominator, which is \((x + 2)(x + 1)\). Rewriting the first term with this denominator:

\[
\frac{{x - 1}}{{x + 2}} = \frac{{(x - 1)(x + 1)}}{{(x + 2)(x + 1)}}
\]

Now the expression becomes:

\[
\lim_{{x \to -2}} \left( \frac{{(x - 1)(x + 1) - (x^2 + x + 1)}}{{(x + 2)(x + 1)}} \right)
\]

### Step 4: Simplify the Numerator

Expand the terms in the numerator:

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

Now subtract the second term:

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

So the expression becomes:

\[
\lim_{{x \to -2}} \frac{{-x - 2}}{{(x + 2)(x + 1)}}
\]

### Step 5: Cancel and Take the Limit

The \(-x - 2\) in the numerator simplifies to \( -(x + 2) \), so we can cancel out the \((x + 2)\) from the numerator and denominator:

\[
\lim_{{x \to -2}} \frac{{-(x + 2)}}{{(x + 2)(x + 1)}} = \lim_{{x \to -2}} \frac{{-1}}{{x + 1}}
\]

Now, substitute \(x = -2\) into the remaining expression:

\[
\frac{{-1}}{{-2 + 1}} = \frac{{-1}}{{-1}} = 1
\]

### Final Answer:

The limit is \( \boxed{1} \).

Would you like further details on any step or another related explanation?

Here are 5 related questions you might find useful:
1. How do you factor a quadratic expression like \(x^2 + 3x + 2\)?
2. What are indeterminate forms in limits, and how are they resolved?
3. How does combining fractions with different denominators work in algebra?
4. Can you always cancel terms in a fraction when they appear in both the numerator and denominator?
5. What are some common techniques for solving limits that result in indeterminate forms?

**Tip**: Factoring is a powerful tool for simplifying limits, especially when encountering indeterminate forms like \(\frac{0}{0}\).

lim (x -> -2) [(x - 1) / (x + 2) - (x^2 + x + 1) / (x^2 + 3x + 2)]

Solve the limit lim (x -> -2) [(x - 1) / (x + 2) - (x^2 + x + 1) / (x^2 + 3x + 2)] using factoring and limit laws. This example demonstrates how to resolve indeterminate forms by simplifying the expression. Ideal for students in Grades 11-12 studying limits and algebraic manipulation.

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