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We are given a problem about population change modeled by the function \( P(t) = 5.4^t \). The task is to determine which of the given statements are always true. Let's examine each statement.

### Statement (i):
\[
P'(t) = \lim_{h \to 0} \frac{5.4^h(5.4^t - 1)}{h}
\]
This is the definition of the derivative of \( P(t) \) at time \( t \), using the limit form. To confirm this, we need to find \( P'(t) \).

Given \( P(t) = 5.4^t \), we apply the general rule for differentiating exponential functions:
\[
P'(t) = \frac{d}{dt} \left( 5.4^t \right) = 5.4^t \ln(5.4)
\]
Thus, the derivative of \( P(t) \) is \( P'(t) = 5.4^t \ln(5.4) \). Therefore, the given limit expression is consistent with this derivative formula. So, **(i) is true**.

---

### Statement (ii):
\[
P'(1) \text{ is less than the average rate of change of } P \text{ between 0 and 1.}
\]
First, we compute \( P'(1) \). Using the formula from statement (i):
\[
P'(1) = 5.4^1 \ln(5.4) = 5.4 \ln(5.4)
\]
Now, the average rate of change of \( P(t) \) between \( t = 0 \) and \( t = 1 \) is:
\[
\text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0} = P(1) - P(0)
\]
Since \( P(1) = 5.4^1 = 5.4 \) and \( P(0) = 5.4^0 = 1 \), the average rate of change is:
\[
5.4 - 1 = 4.4
\]
Now, we compare \( P'(1) = 5.4 \ln(5.4) \) with the average rate of change, \( 4.4 \).

The value of \( \ln(5.4) \) is approximately \( 1.686 \), so:
\[
P'(1) = 5.4 \times 1.686 \approx 9.11
\]
Since \( 9.11 \) is greater than \( 4.4 \), **statement (ii) is false** because \( P'(1) \) is not less than the average rate of change.

---

### Statement (iii):
\[
\text{The average rate of change between 0 and 1 is 5.4.}
\]
We already calculated the average rate of change as \( 4.4 \), so **(iii) is false**.

---

### Conclusion:
From the analysis above, only **statement (i) is true**. Therefore, the correct answer is:

\[
\boxed{A}
\]

Would you like any further clarification or details on these steps?

---

### Related Questions:
1. How is the derivative of an exponential function with a base other than \( e \) calculated?
2. Why is the average rate of change of a function between two points not the same as the instantaneous rate of change?
3. Can you explain how the natural logarithm \( \ln \) plays a role in differentiating exponential functions?
4. What is the significance of the limit definition of the derivative in calculus?
5. How can we interpret the derivative of a population growth function in real-world terms?

### Tip:
For any exponential function \( a^t \), the derivative is \( a^t \ln(a) \), where \( \ln(a) \) is the natural logarithm of the base.

A population changes according to the formula P(t) = 5.4^t. Which of the following statements are always true?
(i) P'(t) = lim(h→0) (5.4^h(5.4^t − 1))/h
(ii) P'(1) is less than the average rate of change of P between 0 and 1.
(iii) The average rate of change between 0 and 1 is 5.4.
Options: (A) (i) only (B) (i) and (ii) only (C) (i) and (iii) only (D) none of them (E) (ii) and (iii) only (F) all of them (G) (iii) only (H) (ii) only

Analyze the population change modeled by P(t) = 5.4^t, including derivative calculations and comparisons of instantaneous and average rates of change. This problem covers key concepts in calculus and exponential functions.

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