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### Problem 1:
We are given the function \( f(x) = \sin \left( \frac{2\pi}{x} \right) \) and a table showing several pairs of inputs and outputs for the function. The goal is to determine what we can conclude about \( \lim_{x \to 0} \sin \left( \frac{2\pi}{x} \right) \) based solely on this table.

The table provides the following information:

| \(x\)  | -1   | -0.1 | -0.01 | -0.001 | 0    | 0.001 | 0.01  | 0.1   | 1     |
|-------|------|------|-------|--------|------|-------|-------|-------|-------|
| \(y = \sin\left( \frac{2\pi}{x} \right)\) | 0    | 0    | 0     | 0      | DNE  | 0     | 0     | 0     | 0     |

The table shows that for values of \( x \neq 0 \), the output \( \sin \left( \frac{2\pi}{x} \right) \) is 0. However, when \( x = 0 \), the function is undefined, denoted by "DNE" (Does Not Exist).

#### Interpretation:
- For nonzero values of \( x \), the function output remains at 0.
- The function is undefined at \( x = 0 \).
- We cannot conclusively determine from the table whether the limit as \( x \to 0 \) exists based solely on the values given.

#### Answer to the question:
From the table alone, we can conclude that \( \lim_{x \to 0} \sin \left( \frac{2\pi}{x} \right) \) is likely to be 0, because the outputs approach 0 as \( x \) approaches 0. Thus, the correct choice from the given options is **(a)**: 
\[
\lim_{x \to 0} \sin \left( \frac{2\pi}{x} \right) = 0.
\]

### Problem 2:
We are asked to determine the actual limit \( \lim_{x \to 0} \sin \left( \frac{2\pi}{x} \right) \).

#### Explanation:
The function \( \sin \left( \frac{2\pi}{x} \right) \) oscillates increasingly rapidly as \( x \) approaches 0. This is because as \( x \) gets smaller, \( \frac{2\pi}{x} \) becomes large, and the sine function oscillates between -1 and 1 very quickly. Since this oscillation does not settle at a particular value, the limit does **not exist**. Therefore, mathematically:
\[
\lim_{x \to 0} \sin \left( \frac{2\pi}{x} \right) \text{ does not exist (DNE)}.
\]

#### Conclusion for Problem 2:
The correct conclusion is that the limit does not exist due to the oscillatory nature of the sine function as \( x \to 0 \).

### Do you want more details or have any questions?

#### 5 Related Questions:
1. What is the general behavior of oscillatory functions as their argument tends to infinity or zero?
2. How can you determine the limit of an oscillatory function without relying solely on a table?
3. What is the difference between limits that approach infinity and those that involve oscillations?
4. Can a limit exist if a function is undefined at a point? Why or why not?
5. How do we rigorously prove that \( \lim_{x \to 0} \sin \left( \frac{2\pi}{x} \right) \) does not exist?

#### Tip:
Always remember that oscillations near a point may cause a limit to not exist, even if the function has predictable outputs away from that point.

Shown are several pairs of inputs and outputs for the function f(x) = sin(2π/x). From the information in this table alone, what can you determine about lim(x→0) sin(2π/x)? Circle the correct answer below:

a. lim(x→0) sin(2π/x) = 0
b. lim(x→0) sin(2π/x) DNE
c. lim(x→0) sin(2π/x) must be close to 0
d. nothing about lim(x→0) sin(2π/x) can be determined from the table alone.

This problem explores the behavior of the function sin(2π/x) as x approaches 0 using a table of values. The problem emphasizes the concept of limits, oscillatory behavior of trigonometric functions, and how to infer the existence of limits from a data table. Suitable for advanced high school or college calculus students.

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