We are given that $f$ is integrable on $[0, x]$ for every $x > 0$, and that $\lim_{x \to \infty} f(x) = a$. Our goal is to prove that

$\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t) \, dt = a.$

Step 1: Split the integral

To apply the hint, choose $N > 0$ and for $x > N$, write $x = N + y$, where $y > 0$. Now split the integral $\int_0^x f(t) \, dt$ into two parts:

$\int_0^x f(t) \, dt = \int_0^N f(t) \, dt + \int_N^x f(t) \, dt.$

So, for $x = N + y$, we have:

$\int_0^x f(t) \, dt = \int_0^N f(t) \, dt + \int_N^{N+y} f(t) \, dt.$

Step 2: Analyze the behavior as $x \to \infty$

Now, we divide both sides of the expression by $x = N + y$:

$\frac{1}{x} \int_0^x f(t) \, dt = \frac{1}{x} \left( \int_0^N f(t) \, dt + \int_N^{N+y} f(t) \, dt \right).$

Let's focus on each term separately:

1. The first term: $\frac{1}{x} \int_0^N f(t) \, dt.$ Since $N$ is fixed and does not depend on $x$, as $x \to \infty$, this term tends to $0$ because $x$ grows without bound, while the integral remains fixed.

2. The second term: $\frac{1}{x} \int_N^{N+y} f(t) \, dt.$ Since $f(x) \to a$ as $x \to \infty$, for sufficiently large $x$, the function $f(t)$ becomes arbitrarily close to $a$ on the interval $[N, N + y]$. So, for large $x$, we can approximate:

   $\int_N^{N+y} f(t) \, dt \approx \int_N^{N+y} a \, dt = a \cdot y.$

   Therefore, the second term becomes:

   $\frac{1}{x} \int_N^{N+y} f(t) \, dt \approx \frac{a \cdot y}{x}.$

   Since $x = N + y$, this simplifies to:

   $\frac{a \cdot y}{x} = \frac{a \cdot y}{N + y}.$

   As $x \to \infty$, we have $y \to \infty$, and thus:

   $\frac{a \cdot y}{N + y} \to a.$

Step 3: Conclusion

Combining the results from the two terms, we find that as $x \to \infty$:

$\frac{1}{x} \int_0^x f(t) \, dt \to a.$

Thus,

$\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t) \, dt = a.$

Further Exploration

1. How does the result change if $\lim_{x \to \infty} f(x)$ does not exist?
2. How would this proof differ if we used the concept of Riemann integrability in more detail?
3. What if $f(x)$ is not continuous but still converges to a limit at infinity?
4. How does this result relate to the concept of the average value of a function?
5. Can you think of situations where this result might be applied in real-world problems, such as in physics or economics?

Tip: This proof uses the idea that if a function converges to a constant $a$ at infinity, then the "tail" of the integral becomes approximately $a \times y$, which makes it possible to find the limit of the average value of the function.