The image contains a series of limit problems labeled from a to g. Let's go through each limit one by one.

a. $\lim_{x \to 0} \frac{3x \sin 4x}{\cos x - 1}$

To solve this, we can use L'Hopital's Rule since both the numerator and denominator tend to 0 as $x \to 0$.

b. $\lim_{x \to \infty} \left( \ln(\sqrt{x}) \right)^{\frac{2}{x}}$

This can be rewritten as: $\lim_{x \to \infty} \left( \frac{1}{2} \ln(x) \right)^{\frac{2}{x}}$ Now, evaluate the limit as $x \to \infty$.

c. $\lim_{x \to 0^+} x^{\tan x}$

This requires us to first find the behavior of $\tan x$ as $x \to 0^+$ and use properties of limits of exponential functions.

d. $\lim_{x \to \frac{\pi}{2}} (1 + \cos x)^{\tan x}$

As $x \to \frac{\pi}{2}$, $\cos x \to 0$ and $\tan x \to \infty$. We need to carefully handle the indeterminate form.

e. $\lim_{x \to 0} \frac{\sin x + e^x}{x^{2021}}$

This is a higher-order limit problem, where we analyze the behavior of both the numerator and the denominator as $x \to 0$.

f. $\lim_{x \to 0} \left[ \frac{\ln(1 + 3x^2)}{x(e^x - 1)} + \left(1 - \frac{1}{x}\right)^{2x} \right]$

This problem involves logarithmic limits and binomial expansions for the second part.

g. $\lim_{x \to 0^+} x \ln(e^{2x} - 1)$

Use logarithmic properties and the fact that $e^{2x} - 1$ can be approximated by $2x$ as $x \to 0^+$.

Would you like to go over any of these limits in more detail, or would you like a step-by-step solution for a specific part?

Here are 5 related questions to consider:

1. How do you apply L'Hopital's Rule in limit calculations?
2. What is the behavior of $\ln(x)$ as $x \to \infty$?
3. How do exponential and logarithmic functions affect limits?
4. How can you handle limits with powers, such as $x^{\tan x}$?
5. What are common techniques for solving limits involving trigonometric functions?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\infty/\infty$ to decide whether L'Hopital's Rule applies!