Let's solve each limit problem step-by-step.

Problem (a)

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

To solve this limit, we can use the approximation of binomial expansions for small $x$ near zero or apply a rationalization technique. Here, rationalizing both the numerator and the denominator by using Taylor series expansion is a reliable approach.

Step 1: Expand the Numerator and Denominator Using Taylor Series

For small $x$, we can use the Taylor expansions around $x = 0$:

1. $\sqrt{1 + x} \approx 1 + \frac{x}{2}$
2. $\sqrt[3]{1 + x} \approx 1 + \frac{x}{3}$

Thus, $\sqrt{1 + x} - 1 \approx \frac{x}{2}$ and $\sqrt[3]{1 + x} - 1 \approx \frac{x}{3}$

Step 2: Substitute These Approximations into the Limit

The limit now becomes $\lim_{x \to 0} \frac{\frac{x}{2}}{\frac{x}{3}}$

Step 3: Simplify the Fraction

$= \lim_{x \to 0} \frac{x}{2} \cdot \frac{3}{x} = \frac{3}{2}$

So, the answer is: $\boxed{\frac{3}{2}}$

Problem (b)

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

This limit appears to be of an indeterminate form of the type $1^\infty$. To handle this, we can take the natural logarithm and use $e$-exponential form.

Step 1: Take the Natural Logarithm of the Expression

Let $y = \left( \frac{x + 1}{2x + 1} \right)^{x^2}$ Then, $\ln y = x^2 \ln \left( \frac{x + 1}{2x + 1} \right)$

Step 2: Simplify the Logarithmic Term Inside the Limit

As $x \to \infty$, we can approximate the argument of the logarithm by observing that: $\frac{x + 1}{2x + 1} = \frac{x(1 + \frac{1}{x})}{2x(1 + \frac{1}{2x})} = \frac{1 + \frac{1}{x}}{2 + \frac{1}{x}} \approx \frac{1}{2} \text{ as } x \to \infty.$

Thus, $\ln \left( \frac{x + 1}{2x + 1} \right) \approx \ln \frac{1}{2} = -\ln 2.$

So, we rewrite the expression as: $\ln y \approx x^2 (-\ln 2) = -x^2 \ln 2.$

Step 3: Take the Limit

Now, take the limit as $x \to \infty$: $\lim_{x \to \infty} \ln y = \lim_{x \to \infty} -x^2 \ln 2 = -\infty.$

Thus, $y = e^{-\infty} = 0.$

So, the answer is: $\boxed{0}$

Summary of Solutions

1. For (a), $\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} = \frac{3}{2}$.
2. For (b), $\lim_{x \to \infty} \left( \frac{x + 1}{2x + 1} \right)^{x^2} = 0$.

Would you like further details on any of these steps, or do you have additional questions?

Here are some related questions to deepen your understanding:

1. How can Taylor series expansions help in solving limits involving square and cube roots?
2. What are common indeterminate forms, and how can they be addressed in limit problems?
3. When can you approximate a fraction like $\frac{x + 1}{2x + 1}$ for large values of $x$?
4. How does the natural logarithm function simplify solving limits of the form $1^\infty$?
5. Can you explain why $e^{-\infty}$ results in $0$ in the context of exponential decay?

Tip: When dealing with limits that produce indeterminate forms like $1^\infty$, taking the natural logarithm is often a powerful strategy to simplify the expression.