Here are the problems from the image (translated as needed) with corresponding solutions:

1. $\lim_{n \to \infty} \frac{\sqrt[3]{n^3 + 2n - 1}}{n + 1}$

Solution:

Factor $n$ from the terms inside the cube root and the denominator: $\frac{\sqrt[3]{n^3(1 + \frac{2}{n^2} - \frac{1}{n^3})}}{n(1 + \frac{1}{n})}$ $= \frac{n \cdot \sqrt[3]{1 + \frac{2}{n^2} - \frac{1}{n^3}}}{n(1 + \frac{1}{n})}.$ As $n \to \infty$, $\sqrt[3]{1 + \frac{2}{n^2} - \frac{1}{n^3}} \to 1$ and $(1 + \frac{1}{n}) \to 1$: $\lim_{n \to \infty} = \frac{1}{1} = 1.$

2. $\lim_{x \to 1} \frac{x^2 - 3x + 2}{x^3 - 6x^2 + 11x - 6}$

Solution:

Factor both numerator and denominator: $\frac{(x - 1)(x - 2)}{(x - 1)(x - 2)(x - 3)}.$ Cancel $(x - 1)(x - 2)$, leaving: $\frac{1}{x - 3}.$ At $x = 1$, $x - 3 = -2$, so: $\lim_{x \to 1} = -\frac{1}{2}.$

3. $\lim_{x \to 0} \frac{1 - \cos 4x}{6x^2}$

Solution:

Using the small-angle approximation $\cos 4x \approx 1 - \frac{(4x)^2}{2}$, we have: $\frac{1 - \cos 4x}{6x^2} \approx \frac{\frac{16x^2}{2}}{6x^2} = \frac{8}{6} = \frac{4}{3}.$

4. $\lim_{x \to 0} \frac{e^{2x} - 1}{3x}$

Solution:

Using the Taylor expansion $e^{2x} \approx 1 + 2x$, we get: $\frac{e^{2x} - 1}{3x} \approx \frac{2x}{3x} = \frac{2}{3}.$

5. $\lim_{x \to 2} \left( \frac{x + 1}{x - 2} \right)^{2x + 1}$

Solution:

Let $y = \left( \frac{x + 1}{x - 2} \right)^{2x + 1}$ and take the logarithm: $\ln y = (2x + 1) \ln \left( \frac{x + 1}{x - 2} \right).$ As $x \to 2$, $2x + 1 \to 5$ and $\ln \left( \frac{x + 1}{x - 2} \right) \to \infty$. The limit diverges to $\infty$.

6. $\lim_{x \to 0} \frac{\sqrt[3]{1 + x} - 1}{x^2}$

Solution:

Expand $\sqrt[3]{1 + x}$ using the binomial theorem: $\sqrt[3]{1 + x} \approx 1 + \frac{x}{3}.$ The numerator becomes: $\sqrt[3]{1 + x} - 1 \approx \frac{x}{3}.$ Thus: $\frac{\sqrt[3]{1 + x} - 1}{x^2} \approx \frac{\frac{x}{3}}{x^2} = \frac{1}{3x}.$ As $x \to 0$, the limit diverges to $\infty$.

7. $\lim_{n \to \infty} \frac{1 + 3 + 5 + \dots + (2n - 1)}{n + 1} \cdot \frac{2n + 1}{2}$

Solution:

The sum of the first $n$ odd numbers is: $1 + 3 + 5 + \dots + (2n - 1) = n^2.$ Substitute this into the expression: $\frac{n^2}{n + 1} \cdot \frac{2n + 1}{2}.$ Factor out $n^2$ in the numerator and simplify: $\frac{n^2}{n + 1} \cdot \frac{2n + 1}{2} \approx \frac{n^2}{n} \cdot n = \frac{n}{2}.$ As $n \to \infty$, the limit grows without bound ($\infty$).

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Relative Questions:

1. What is the significance of L'Hôpital's Rule in solving indeterminate limits?
2. How does Taylor series approximation simplify solving limits involving exponential or trigonometric functions?
3. Why does the binomial theorem help when expanding fractional powers near zero?
4. Can the logarithmic transformation simplify more complex limit problems involving exponents?
5. How does divergence behave differently for polynomial versus transcendental growth rates?

Tip: When dealing with $\infty$ or $0/0$ limits, always check if L'Hôpital's Rule or series expansions simplify the problem before direct substitution.