The given problem describes a population function $P(t) = 5.4^t$ and asks which of three statements about the population's rate of change are always true.

Let's analyze each statement:

1. Statement (i): $P'(t) = \lim_{h \to 0} \frac{5.4^h (5.4^t - 1)}{h}$

   • This appears to represent the derivative of the population function $P(t) = 5.4^t$, where $P'(t)$ would be calculated using differentiation rules. Using exponential differentiation: $P'(t) = \frac{d}{dt} 5.4^t = 5.4^t \ln(5.4)$ This limit representation aligns with the derivative formula, meaning statement (i) is true.

2. Statement (ii): $P'(1)$ is less than the average rate of change of $P$ between 0 and 1.

   • First, we compute $P'(1)$: $P'(1) = 5.4^1 \ln(5.4) = 5.4 \ln(5.4)$
   • Next, we compute the average rate of change between $t = 0$ and $t = 1$: $\text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0} = P(1) - P(0) = 5.4^1 - 5.4^0 = 5.4 - 1 = 4.4$
   • Now we compare $P'(1) \approx 5.4 \times 1.686 \approx 9.11$ with the average rate of change (4.4). Since $9.11 > 4.4$, statement (ii) is false.

3. Statement (iii): The average rate of change between 0 and 1 is 5.4.

   • We already computed the average rate of change between 0 and 1, which is 4.4, not 5.4. Therefore, statement (iii) is false.

Conclusion:

• Statement (i) is true.
• Statements (ii) and (iii) are false.

The correct answer is $\boxed{A}$ (i) only.

Do you have any questions or would you like further details? Here are some related questions to explore:

1. How do you differentiate exponential functions in general?
2. What is the difference between the derivative and the average rate of change?
3. Why is the natural logarithm involved in the derivative of exponential functions?
4. Can the rate of change of an exponential function ever be less than the average rate of change?
5. What does the derivative of a function represent in practical terms?

Tip: When comparing a function’s derivative to its average rate of change, remember that the derivative is the instantaneous rate, while the average rate covers a broader interval.