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### Problem 3:
\[
\lim_{x \to +\infty} \left(\frac{1 + 1/x}{x}\right)
\]

### Solution for Problem 3:
As \( x \to +\infty \), \( \frac{1}{x} \to 0 \). Therefore, \( 1 + \frac{1}{x} \to 1 \).
\[
\frac{1 + \frac{1}{x}}{x} \approx \frac{1}{x} \to 0 \text{ as } x \to +\infty.
\]
**Answer**: 0.

---

### Problem 4:
\[
\lim_{x \to 0^-} \frac{\sin x + \tan x}{x}
\]

### Solution for Problem 4:
As \( x \to 0^- \), both \( \sin x \) and \( \tan x \) can be approximated by \( x \), thus:
\[
\frac{\sin x + \tan x}{x} \approx \frac{x + x}{x} = 2.
\]
**Answer**: 2.

---

### Problem 5:
\[
\lim_{x \to +\infty} \frac{1}{\sqrt{x^2 + x} - x}
\]

### Solution for Problem 5:
Rationalize the denominator:
\[
\frac{1}{\sqrt{x^2 + x} - x} \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x} = \frac{\sqrt{x^2 + x} + x}{x}.
\]
Then, as \( x \to +\infty \), this simplifies to 1.
**Answer**: 1.

---

### Problem 6:
\[
\lim_{x \to 0^+} \frac{1}{x^{1/2}}
\]

### Solution for Problem 6:
As \( x \to 0^+ \), \( x^{1/2} \to 0 \), making \(\frac{1}{x^{1/2}}\) diverge to \(+\infty\).
**Answer**: \( +\infty \).

---

### Problem 7:
\[
\lim_{x \to 0} ((\sin x)^2 + (\cos x)^2)
\]

### Solution for Problem 7:
Using the Pythagorean identity:
\[
(\sin x)^2 + (\cos x)^2 = 1.
\]
**Answer**: 1.

---

### Problem 8:
\[
\lim_{x \to +\infty} \left(\frac{1 + 1/x}{x} - 2x\right)
\]

### Solution for Problem 8:
First term \(\to 0\)

Solve the limit problems: 
1. \( \lim_{x \to +\infty} \frac{1 + 1/x}{x} \)
2. \( \lim_{x \to 0^-} \frac{\sin x + \tan x}{x} \)
3. \( \lim_{x \to +\infty} \frac{1}{\sqrt{x^2 + x} - x} \)
4. \( \lim_{x \to 0^+} \frac{1}{x^{1/2}} \)
5. \( \lim_{x \to 0} ((\sin x)^2 + (\cos x)^2) \)
6. \( \lim_{x \to +\infty} \left(\frac{1 + 1/x}{x} - 2x\right) \)

Explore solutions for various limit problems including indeterminate forms, rationalizing denominators, and using L'Hôpital's Rule. Problems involve limits as x approaches infinity and zero, Pythagorean identities, and exponential functions. Perfect for college-level calculus students.

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