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The problem asks to analyze the population growth given by the function \( P(t) = 5.4^t \), and it provides three statements, asking which are always true.

Let's break down each statement one by one:

### (i) \( P'(t) = \lim_{h \to 0} \frac{5.4^h(5.4^t - 1)}{h} \)
This is a limit definition for the derivative of the function \( P(t) = 5.4^t \). To confirm if this is correct, recall that the derivative of \( a^t \) (for some constant \( a \)) is:
\[
P'(t) = a^t \ln(a)
\]
So for \( P(t) = 5.4^t \), the derivative should be:
\[
P'(t) = 5.4^t \ln(5.4)
\]
Without calculating the limit explicitly, this matches the general form of the derivative for an exponential function. Thus, **statement (i)** appears to be true.

### (ii) \( P'(1) \) is less than the average rate of change of \( P \) between 0 and 1.
To check this, we first calculate \( P'(1) \). Using the formula for the derivative:
\[
P'(1) = 5.4^1 \ln(5.4)
\]
Next, compute the average rate of change of \( P(t) \) from \( t = 0 \) to \( t = 1 \):
\[
\text{Average rate of change} = \frac{P(1) - P(0)}{1 - 0} = \frac{5.4^1 - 5.4^0}{1} = 5.4 - 1 = 4.4
\]
Now, let's estimate \( P'(1) \):
\[
P'(1) = 5.4 \ln(5.4)
\]
Since \( \ln(5.4) \approx 1.686 \), we get:
\[
P'(1) \approx 5.4 \times 1.686 = 9.11
\]
So, \( P'(1) \approx 9.11 \), which is greater than the average rate of change \( 4.4 \). Therefore, **statement (ii)** is false.

### (iii) The average rate of change between 0 and 1 is 5.4.
From the previous calculation, we know that the average rate of change between 0 and 1 is \( 4.4 \), not 5.4. Hence, **statement (iii)** is also false.

### Conclusion:
Only **statement (i)** is true. The correct answer is (A) "i only."

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Do you need further details on any part of this problem?

Here are five related questions you might find helpful:
1. How do you find the derivative of any exponential function \( a^t \)?
2. What is the significance of the natural logarithm \( \ln(a) \) in the derivative of exponential functions?
3. How can you estimate the derivative using the limit definition?
4. How do you compare instantaneous rate of change with average rate of change?
5. What are some real-world applications of exponential growth models like \( P(t) = 5.4^t \)?

**Tip:** When comparing the average rate of change and instantaneous rate of change, always remember the instantaneous rate (derivative) gives the rate at one point, while the average rate gives the overall change across an interval.

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 a population growth model given by P(t) = 5.4^t. Learn how to compute the derivative using the limit definition, compare instantaneous and average rates of change, and verify which statements about the function are true.

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