Determine the sollowing limits and explain your reasoning. a) $$ \lim _{x \rightarrow-3} \frac{x^{2}-4 x+3}{x^{2}+3 x+2} $$

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To determine the limit

$$
\lim _{x \rightarrow -3} \frac{x^{2}-4x+3}{x^{2}+3x+2},
$$

we will follow these steps:

1. **Substitute the value of $$x$$**: First, we will substitute $$x = -3$$ directly into the expression to see if we can evaluate the limit directly.

2. **Factor the numerator and denominator**: If direct substitution results in an indeterminate form (like $$\frac{0}{0}$$), we will factor both the numerator and the denominator to simplify the expression.

3. **Re-evaluate the limit**: After simplification, we will re-evaluate the limit.

Let's start with step 1:

### Step 1: Direct Substitution

Substituting $$x = -3$$:

- Numerator: 
$$
(-3)^{2} - 4(-3) + 3 = 9 + 12 + 3 = 24.
$$

- Denominator: 
$$
(-3)^{2} + 3(-3) + 2 = 9 - 9 + 2 = 2.
$$

So, we have:

$$
\frac{24}{2} = 12.
$$

Since we did not get an indeterminate form, we can conclude that:

$$
\lim _{x \rightarrow -3} \frac{x^{2}-4x+3}{x^{2}+3x+2} = 12.
$$

### Conclusion

The limit is

$$
\boxed{12}.
$$

This result was obtained by direct substitution, confirming that the function is continuous at $$x = -3$$.
The limit is 12.

Determine the sollowing limits and explain your reasoning. a) \( \lim _{x \rightarrow-3} \frac{x^{2}-4 x+3}{x^{2}+3 x+2} \)

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