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Let's solve the calculus problems one by one from the image you've uploaded:

### Problem 1: 
\[
\lim_{x \to -1} \frac{x^2 + 3x + 2}{x+1}
\]
We can first simplify the given expression by factoring the numerator:
\[
x^2 + 3x + 2 = (x + 1)(x + 2)
\]
Thus, the expression becomes:
\[
\frac{(x + 1)(x + 2)}{x + 1}
\]
For \(x \neq -1\), we can cancel the \(x + 1\) terms:
\[
\frac{(x + 1)(x + 2)}{x + 1} = x + 2
\]
Now, we can directly substitute \(x = -1\) into the simplified expression:
\[
x + 2 = -1 + 2 = 1
\]
Therefore, 
\[
\lim_{x \to -1} \frac{x^2 + 3x + 2}{x+1} = 1
\]

---

### Problem 2:
\[
\lim_{x \to 0} f(x), \quad \text{where } f(x) = \begin{cases} 
3x + 1 & \text{if } x \leq 0 \\
2 & \text{if } x > 0 
\end{cases}
\]

We need to check the left-hand limit and the right-hand limit separately:
- For the **left-hand limit** \(x \to 0^{-}\), we use \(f(x) = 3x + 1\):
  \[
  \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} (3x + 1) = 1
  \]

- For the **right-hand limit** \(x \to 0^{+}\), we use \(f(x) = 2\):
  \[
  \lim_{x \to 0^{+}} f(x) = 2
  \]

Since the left-hand limit and right-hand limit are not equal, the limit **does not exist**:
\[
\lim_{x \to 0} f(x) \text{ does not exist.}
\]

---

### Problem 3:
\[
\lim_{x \to -1} \frac{x^2}{\sqrt{x+2}}
\]
First, substitute \(x = -1\) directly:
\[
\frac{(-1)^2}{\sqrt{-1 + 2}} = \frac{1}{\sqrt{1}} = 1
\]
Thus, the limit is:
\[
\lim_{x \to -1} \frac{x^2}{\sqrt{x+2}} = 1
\]

---

### Problem 4 (not shown):
We will skip this problem as it is not visible in the uploaded image.

---

### Problem 5:
\[
\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}}
\]
We can divide both the numerator and the denominator by \(x\):
\[
\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}} = \lim_{x \to \infty} \frac{x(3 - \frac{2}{x})}{x\sqrt{1 - \frac{10}{x^2}}}
\]
This simplifies to:
\[
\lim_{x \to \infty} \frac{3 - \frac{2}{x}}{\sqrt{1 - \frac{10}{x^2}}}
\]
As \(x \to \infty\), \(\frac{2}{x} \to 0\) and \(\frac{10}{x^2} \to 0\), so the expression becomes:
\[
\frac{3 - 0}{\sqrt{1 - 0}} = \frac{3}{1} = 3
\]
Thus, the limit is:
\[
\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}} = 3
\]

---

### Summary:
1. \(\lim_{x \to -1} \frac{x^2 + 3x + 2}{x+1} = 1\)
2. \(\lim_{x \to 0} f(x)\) does not exist.
3. \(\lim_{x \to -1} \frac{x^2}{\sqrt{x+2}} = 1\)
5. \(\lim_{x \to \infty} \frac{3x - 2}{\sqrt{x^2 - 10}} = 3\)

Do you have any questions, or would you like more details?

### Related Questions:
1. What is the general process for solving limits using factoring?
2. How do piecewise functions affect the existence of limits?
3. What are the strategies for evaluating limits as \(x \to \infty\)?
4. How can you recognize when a limit does not exist from a graph?
5. How does L'Hôpital's rule apply to evaluating difficult limits?

### Tip:
Always simplify the expression first, if possible, before trying to directly evaluate a limit. This often helps avoid indeterminate forms.

Math 1341 Calculus 1 Homework Chapter 1.3: Limit problems including factoring, piecewise functions, and limits at infinity.

Learn how to solve calculus problems involving limits using factoring, piecewise functions, and limits at infinity. This problem set covers core concepts from Calculus 1, including factoring quadratics, determining limits from graphs, and analyzing piecewise functions.

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