I remember having a hard time solving this in my first year of college. This is really impressive.

So I have this problem here: given that for positive integers m and n, 0 ≤ x[m + n] ≤ x[m] + x[n], I need to prove that the limit of x[n]/n exists. Hmm, okay, let's see how to approach this.

…

Alternative approach: maybe I can show that a_n is convergent by showing that it's a Cauchy sequence.

But that might be complicated.

Alternatively, perhaps I can use the subadditivity to bound a_n.

Wait, let's consider the following:

Since x is subadditive, x[m + n] ≤ x[m] + x[n].

Let me set m = n, then x[2n] ≤ x[n] + x[n] = 2x[n].

Therefore, x[2n]/(2n) ≤ x[n]/n.

So, a_{2n} ≤ a_n.

That means that a_{2n} ≤ a_n.

Similarly, x[3n] ≤ x[2n] + x[n] ≤ 2x[n] + x[n] = 3x[n].

So, x[3n]/(3n) ≤ x[n]/n.

Similarly, x[4n]/(4n) ≤ x[2n]/(2n) ≤ x[n]/n.

It seems like a_{kn} ≤ a_n for any integer k ≥ 1.

This suggests that as n increases, a_n tends to decrease or stay the same.

But wait, in the example where x[n] = n, a_n = 1 for all n, so it's constant.

In the example where x[n] = sqrt(n), a_n = 1/sqrt(n), which decreases to 0.

So, in some cases, a_n is decreasing, and in some cases, it's constant.

But in general, a_n is bounded below by 0 and is decreasing along powers of 2, for example.

Wait, from x[2n]/(2n) ≤ x[n]/n, and x[4n]/(4n) ≤ x[2n]/(2n), and so on, so along the sequence n, 2n, 4n, 8n, etc., a_n is decreasing.

Maybe I can use this to show that a_n has a limit.

Let me consider n = 2^k, for integer k.

Let b_k = x[2^k]/2^k.

Then, from x[2^{k+1}] ≤ x[2^k] + x[2^k] = 2x[2^k], so b_{k+1} = x[2^{k+1}]/2^{k+1} ≤ (2x[2^k])/2^{k+1} = x[2^k]/2^k = b_k.

So, b_{k+1} ≤ b_k, meaning that b_k is a decreasing sequence.

Since b_k ≥ 0, it must converge to some limit, say L.

Now, I need to show that x[n]/n approaches L as n approaches infinity.

Given that b_k is decreasing and converges to L, and for general n, a_n is bounded below by L, and perhaps approaches L.

Wait, more carefully:

For any n, there exists k such that 2^{k-1} < n ≤ 2^k.

Then, x[n]/n ≤ x[2^k]/2^{k-1} = 2(x[2^k]/2^k) = 2b_k.

Since b_k approaches L, for large k, b_k is close to L.

But I need to show that x[n]/n is close to L.

Wait, perhaps I can bound x[n]/n between b_k and something else.

Given that 2^{k-1} < n ≤ 2^k, then x[n] ≤ x[2^{k-1}] + x[n - 2^{k-1}] ≤ x[2^{k-1}] + x[2^{k-1}] = 2x[2^{k-1}].

Therefore, x[n]/n ≤ (2x[2^{k-1}])/n ≤ (2x[2^{k-1}])/2^{k-1} = 2(x[2^{k-1}]/2^{k-1}) = 2b_{k-1}.

But b_{k-1} approaches L as k increases.

Similarly, since n > 2^{k-1}, x[n]/n > x[2^k]/2^k = b_k.

But b_k ≤ b_{k-1}, and both approach L.

Therefore, x[n]/n is squeezed between b_k and 2b_{k-1}, both of which approach L.

Therefore, x[n]/n approaches L as n approaches infinity.

This suggests that the limit exists and is equal to L.

Therefore, the limit of x[n]/n exists.

Alternatively, if I recall correctly, this is exactly what Fekete's lemma states.

So, in conclusion, the limit of x[n]/n exists.

*Final Answer*

\boxed{\lim_{n \to \infty} \frac{x[n]}{n} \text{ exists}}