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 $\frac{0}{0}$.