The description you provided outlines the Sieve of Eratosthenes, a classic algorithm used to find all prime numbers up to a given limit, \( n \). This algorithm efficiently lists all primes below \( n \) by iteratively marking the multiples of each prime number starting from \( 2 \).

Here's a Python code implementation to perform the Sieve of Eratosthenes:

```python
def sieve_of_eratosthenes(n):
    # Create a boolean array "prime[0..n]" and initialize all entries as True.
    # Entry will finally be False if it is Not a prime, True if it is a prime.
    prime = [True for _ in range(n+1)]
    p = 2
    while (p * p <= n):
        # If prime[p] is not changed, then it is a prime
        if (prime[p] == True):
            # Updating all multiples of p to False starting from p*p
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    
    # Collect all prime numbers
    primes = [p for p in range(2, n) if prime[p]]
    return primes

# Example use
x = 30
print(f"All prime numbers smaller than {x}: {sieve_of_eratosthenes(x)}")
```

This will output:
```
All prime numbers smaller than 30: [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
```

### Algorithm explanation:
1. **Initialization**: A list `prime` is initialized with `True` values indicating all numbers are initially assumed prime except number 0 and 1.
2. **Marking non-primes**: Starting from the first prime number \(2\), mark all its multiples as non-prime. Increment the base prime and repeat till you've considered numbers up to the square root of \( n \).
3. **Collect results**: Extract numbers which remain `True` in the list, indicating these are prime numbers.

### Performance:
The Sieve of Eratosthenes runs in \( O(n \log \log n) \) time complexity, making it highly efficient for large \( n \). This performance is derived from the fact that it marks the multiples of each prime number only, minimizing redundant operations.

The memory required is \( O(n) \), to store the boolean values for each number up to \( n \).