1. Define the function and initial setup:
For every positive integer $n$, let $s(n)$ be the sum of the exponents of $71$ and $97$ in the prime factorization of $n$. For example, $s(2021) = s(43 \cdot 47) = 0$ and $s(488977) = s(71^2 \cdot 97) = 3$. Define $f(n) = (-1)^{s(n)}$.

2. **Sum function $S(n)$:**
Let $S(n)$ denote the sum $\sum_{x = 1}^n f(x)$.

3. Subsets and their properties:
Consider the set $[n] = \{1, 2, \ldots, n\}$. Define subsets:
- $A = \{x \in [n] : 71 \mid x\}$
- $B = \{x \in [n] : 97 \mid x\}$
- $A \cap B = \{x \in [n] : 71 \cdot 97 \mid x\}$

We have:
- $[n] \setminus (A \cup B) = \{x \in [n] : \gcd(x, 71 \cdot 97) = 1\}$, which implies $f(x) = 1$ for all $x \in [n] \setminus (A \cup B)$.

4. **Expression for $S(n)$:**
$$S(n) = \sum_{x \in [n]} f(x) = \left( \sum_{x \in A} f(x) + \sum_{x \in B} f(x) - \sum_{x \in A \cap B} f(x) + \sum_{x \in [n] \setminus (A \cup B)} f(x) \right)$$
Since $f(x) = -1$ for $x \in A \cup B$ and $f(x) = 1$ for $x \in [n] \setminus (A \cup B)$, we get:
$$S(n) = \sum_{x \in [\lfloor \frac{n}{71} \rfloor]} (-f(x)) + \sum_{x \in [\lfloor \frac{n}{97} \rfloor]} (-f(x)) - \sum_{x \in [\lfloor \frac{n}{71 \cdot 97} \rfloor]} f(x) + (n - \lfloor \frac{n}{71} \rfloor - \lfloor \frac{n}{97} \rfloor + \lfloor \frac{n}{71 \cdot 97} \rfloor)$$
Simplifying, we get:
$$S(n) = n - \lfloor \frac{n}{71} \rfloor - \lfloor \frac{n}{97} \rfloor + \lfloor \frac{n}{71 \cdot 97} \rfloor - S(\lfloor \frac{n}{71} \rfloor) - S(\lfloor \frac{n}{97} \rfloor) - S(\lfloor \frac{n}{71 \cdot 97} \rfloor)$$

5. **Define $\lambda$ and $D(n)$:**
Let $\lambda = \frac{(71 - 1)(97 - 1)}{(71 + 1)(97 + 1)}$ and $D(n) = S(n) - \lambda n$.

6. Transform the identity:
The above identity translates to:
$$D(n) + D(\lfloor \frac{n}{71} \rfloor) + D(\lfloor \frac{n}{97} \rfloor) + D(\lfloor \frac{n}{71 \cdot 97} \rfloor) = (1 + \lambda) \left( \left\{ \frac{n}{71} \right\} + \left\{ \frac{n}{97} \right\} \right) - (1 - \lambda) \left\{ \frac{n}{71 \cdot 97} \right\}$$
where $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.

7. **Bound $|D(n)|$:**
It follows that:
$$|D(n)| \leq |D(\lfloor \frac{n}{71} \rfloor)| + |D(\lfloor \frac{n}{97} \rfloor)| + |D(\lfloor \frac{n}{71 \cdot 97} \rfloor)| + 4$$

8. **Choose $r$ and bound $D(n)$:**
Choose any positive real number $r < 1$ such that $\epsilon = 1 - \left( \frac{1}{71^r} + \frac{1}{97^r} + \frac{1}{(71 \cdot 97)^r} \right)$ is positive (e.g., $r = \frac{1}{2}$). Let $M$ denote the number $\max \left( \frac{4}{\epsilon}, \max \{ n^{-r} |D(n)| : 1 \leq n < 71 \cdot 97 \} \right)$.

9. **Induction on $n$:**
Prove by induction on $n$ that $n^{-r} |D(n)| \leq M$ for any $n \geq 1$. The claim is true for $1 \leq n < 71 \cdot 97$. Suppose it is true for all positive integers less than $n$. Then for $n$, we have:
$$|D(n)| \leq \lfloor \frac{n}{71} \rfloor^r M + \lfloor \frac{n}{97} \rfloor^r M + \lfloor \frac{n}{71 \cdot 97} \rfloor^r M + \epsilon M \leq n^r M$$
Thus, the claim is true for all $n \geq 1$.

10. Conclusion:
We have shown that $|S(n) - \lambda n| \leq n^r M$ for some constants $M > 0$ and $r < 1$. It follows that:
$$\lim_{n \to \infty} \frac{S(n)}{n} = \lambda = \frac{(71 - 1)(97 - 1)}{(71 + 1)(97 + 1)} = \frac{20}{21}$$

The final answer is $\boxed{ \frac{20}{21} }$