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You have provided an image with several limit problems. I will solve these one by one, providing step-by-step explanations for each. Let's begin with the first few.

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### 1) \(\lim_{x \to 1} \frac{\sqrt{x^2 + 3x + 1} - 5x}{2x + 1}\)

To solve this, let's first try direct substitution by plugging in \(x = 1\):

\[
\text{Numerator:} \quad \sqrt{1^2 + 3(1) + 1} - 5(1) = \sqrt{1 + 3 + 1} - 5 = \sqrt{5} - 5
\]
\[
\text{Denominator:} \quad 2(1) + 1 = 3
\]

Thus, the limit becomes:
\[
\lim_{x \to 1} \frac{\sqrt{5} - 5}{3}
\]

Since this is a constant, the limit is simply:

\[
\boxed{\frac{\sqrt{5} - 5}{3}}
\]

---

### 2) \(\lim_{x \to \infty} \frac{3x^3 + 5x + 1}{2x^3 + x^2 - 9}\)

We will divide the numerator and the denominator by \(x^3\) (the highest power of \(x\)) to simplify:

\[
\lim_{x \to \infty} \frac{\frac{3x^3}{x^3} + \frac{5x}{x^3} + \frac{1}{x^3}}{\frac{2x^3}{x^3} + \frac{x^2}{x^3} - \frac{9}{x^3}} = \lim_{x \to \infty} \frac{3 + \frac{5}{x^2} + \frac{1}{x^3}}{2 + \frac{1}{x} - \frac{9}{x^3}}
\]

As \(x \to \infty\), all terms with \( \frac{1}{x^n} \) vanish, leaving:

\[
\lim_{x \to \infty} \frac{3}{2} = \boxed{\frac{3}{2}}
\]

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### 3) \(\lim_{x \to 2} \frac{x^3 + 8}{x^2 - 3x - 10}\)

First, factor the denominator:

\[
x^2 - 3x - 10 = (x - 5)(x + 2)
\]

Substitute \(x = 2\) directly:

\[
\text{Numerator:} \quad 2^3 + 8 = 8 + 8 = 16
\]
\[
\text{Denominator:} \quad (2 - 5)(2 + 2) = (-3)(4) = -12
\]

Thus, the limit is:

\[
\lim_{x \to 2} \frac{16}{-12} = \boxed{-\frac{4}{3}}
\]

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### 4) \(\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}\)

This is a classic indeterminate form. We can use the conjugate to simplify:

Multiply numerator and denominator by the conjugate of the numerator:

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

Now cancel \(x - 1\):

\[
\frac{1}{\sqrt{x} + 1}
\]

Substitute \(x = 1\):

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

Thus, the limit is:

\[
\boxed{\frac{1}{2}}
\]

---

Would you like me to continue solving the rest of the problems in the image? Please let me know!

Here are 5 related questions you might find helpful:
1. How do you handle indeterminate forms like \(0/0\)?
2. What are common techniques to simplify complex limit problems?
3. How can the conjugate method help in solving limits involving square roots?
4. What is L'Hopital's rule and when should it be used?
5. How do we approach limits involving infinity in rational functions?

**Tip:** Always check for factorizations and simplifications before applying advanced techniques like L'Hopital's rule.

Solve the limit problems provided in the image.

Explore a detailed solution to multiple limit problems, including indeterminate forms, L'Hopital's Rule, and rational functions. Perfect for undergraduate calculus students.

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