Suppose that f is integrable on [0, x] for every x > 0, and that
limx→∞ f (x) = a. Prove that
lim
x→∞
1
x
∫ x
0
f (t) dt = a .
Hint: For N > 0 (to be chosen later) and x > N , write x = N + y and split the integral as
∫ N +y
0
f (t) dt =
∫ N
0
f (t) dx +
∫ N +y
N
f (t) dt .
Take N large and compare the second integral with a · y

To prove that

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

given that \( f \) is integrable on \([0, x]\) for every \( x > 0 \) and that \( \lim_{x \to \infty} f(x) = a \), we can follow the hint provided and proceed step-by-step.

### Step 1: Split the integral
For \( x > N \), write \( x = N + y \), where \( y > 0 \), and then split the integral as suggested:

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

Thus, we have

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

### Step 2: Deal with the first integral
The first integral \( \int_0^N f(t) \, dt \) is independent of \( x \) and thus becomes negligible as \( x \to \infty \). Specifically,

\[
\lim_{x \to \infty} \frac{1}{x} \int_0^N f(t) \, dt = 0.
\]

### Step 3: Estimate the second integral
Now, consider the second integral:

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

For large \( x \), since \( \lim_{t \to \infty} f(t) = a \), for sufficiently large \( N \), \( f(t) \) is close to \( a \) for all \( t \in [N, x] \). More formally, for any \( \epsilon > 0 \), there exists some \( N_\epsilon \) such that for all \( t > N_\epsilon \), \( |f(t) - a| < \epsilon \).

Thus, for \( x > N_\epsilon \), we can estimate:

\[
\left| \int_N^x f(t) \, dt - \int_N^x a \, dt \right| = \left| \int_N^x (f(t) - a) \, dt \right| \leq \int_N^x |f(t) - a| \, dt.
\]

Since \( |f(t) - a| < \epsilon \) for \( t > N_\epsilon \), we have

\[
\int_N^x |f(t) - a| \, dt \leq \epsilon (x - N) = \epsilon y.
\]

Therefore,

\[
\int_N^x f(t) \, dt = a(x - N) + O(x - N) = a y + O(y).
\]

### Step 4: Divide by \( x \)
Now consider the full expression:

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

From the previous steps, we have:

- \( \frac{1}{x} \int_0^N f(t) \, dt \to 0 \) as \( x \to \infty \),
- \( \frac{1}{x} \int_N^x f(t) \, dt = \frac{a y + O(y)}{x} = \frac{a y}{x} + O\left( \frac{y}{x} \right) \).

Since \( y = x - N \) and as \( x \to \infty \), \( \frac{y}{x} \to 1 \), we conclude that

\[
\frac{1}{x} \int_N^x f(t) \, dt \to a.
\]

### Step 5: Conclusion
Therefore,

\[
\lim_{x \to \infty} \frac{1}{x} \int_0^x f(t) \, dt = 0 + a = a.
\]

Hence, we have shown that:

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

This completes the proof.

---

Would you like more details on any part of the proof, or do you have additional questions?

Here are 5 related questions that might help further your understanding:
1. How do we use the definition of the limit to prove that \( \lim_{x \to \infty} f(x) = a \)?
2. What is the significance of the integral test for convergence in this context?
3. How does the concept of Riemann integrability relate to the problem at hand?
4. What would happen if \( f(x) \) did not tend to \( a \) as \( x \to \infty \)?
5. How does the concept of a "local error" in the integral \( \int_N^x (f(t) - a) \, dt \) help us bound the difference between the two integrals?

Tip: The key idea in these kinds of proofs is breaking the problem into manageable parts and handling the asymptotic behavior of the function carefully.

Suppose that f is integrable on [0, x] for every x > 0, and that limx→∞ f(x) = a. Prove that lim x→∞ 1/x ∫ x 0 f(t) dt = a. Hint: For N > 0 (to be chosen later) and x > N, write x = N + y and split the integral as ∫ N + y 0 f(t) dt = ∫ N 0 f(t) dx + ∫ N + y N f(t) dt. Take N large and compare the second integral with a · y.

This detailed proof shows that for an integrable function f on [0, x] with limx→∞ f(x) = a, the limit of the average value of f from 0 to x approaches a. The proof uses integral calculus, limits, and the squeeze theorem to break down the integral. Suitable for college-level calculus students.

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