The image you have uploaded contains a set of 14 limit problems under the title "Latihan Soal." Let me provide solutions to a few of these problems with detailed steps.

Problem 1:

$\lim_{x \to 0} \frac{\cos x}{x + 1}$

Solution:

1. As $x \to 0$, $\cos x \to \cos 0 = 1$ and $x + 1 \to 1$.

Thus, the limit is: $\lim_{x \to 0} \frac{\cos 0}{0 + 1} = \frac{1}{1} = 1$

Problem 2:

$\lim_{\theta \to \frac{\pi}{2}} \theta \cos \theta$

Solution:

1. As $\theta \to \frac{\pi}{2}$, $\cos \theta \to \cos \frac{\pi}{2} = 0$.

Thus, the limit is: $\lim_{\theta \to \frac{\pi}{2}} \theta \cos \theta = \frac{\pi}{2} \cdot 0 = 0$

Problem 3:

$\lim_{t \to 0} \frac{\cos^2 t}{1 + \sin t}$

Solution:

1. As $t \to 0$, $\cos^2 t \to \cos^2(0) = 1$ and $\sin t \to 0$, so $1 + \sin t \to 1$.

Thus, the limit is: $\lim_{t \to 0} \frac{\cos^2 0}{1 + \sin 0} = \frac{1}{1} = 1$

Problem 4:

$\lim_{x \to 0} \frac{\sin x}{3x \tan x}$

Solution:

1. Use the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{\tan x}{x} = 1$.

Thus, the limit becomes: $\lim_{x \to 0} \frac{\frac{\sin x}{x}}{3 \frac{\tan x}{x}} = \frac{1}{3} = \frac{1}{3}$

Problem 5:

$\lim_{x \to 0} \frac{\sin x}{2x}$

Solution:

1. Again, use $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Thus, the limit becomes: $\lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2}$

Do you want further explanations or solutions to more of these problems? Here are some relative questions for you:

1. How do you evaluate limits when the function involves a trigonometric identity?
2. What are common limit rules used in calculus, especially for trigonometric limits?
3. How does L'Hôpital's Rule apply to limits with indeterminate forms like $0/0$?
4. What is the small-angle approximation for trigonometric functions, and how is it useful in limits?
5. Can limits always be calculated directly, or do you sometimes need additional techniques (like series expansions)?

Tip: When dealing with trigonometric limits, remembering that $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ simplifies many problems.