Question 6
Substitutions for $h$ are used to estimate the limit of $\dfrac{a^h-1}{h}$ as $h\to0$. Which sequence is the most appropriate?
That’s it. Step through the hints below to see why. In 2023, 59.9% of Queensland students chose (A).
Not quite. These start close to 0, but each value is double the one before, so they move away from 0. The best estimate would be the first one, and every substitution after it is worse. In 2023, 16.1% of Queensland students chose (B).
Not quite. This sequence substitutes $h=0$, which gives $\tfrac{0}{0}$, and that is undefined. The steps also stay 1 apart, so the values never close in on 0. In 2023, 15.9% of Queensland students chose (C).
Not quite. These values get bigger, so they move further from 0 every time. They tell you about large values of $h$, not about $h\to0$. In 2023, 7.4% of Queensland students chose (D).
Drawn with $a=2$. A chord to a point above the top of the graph is cut off at the edge.
Each $h$ is half the one before, so the substitutions close in on 0 without ever reaching it. The chords swing in towards the tangent and the gradients climb towards $0.6931$.
One substitution can’t show a trend. Add more and watch how far each $h$ is from 0.
Each $h$ is double the one before, so the substitutions move away from 0. The first chord is the closest to the tangent, and each one after it is further away.
This first value is close to 0, so this chord is close to the tangent. Add more substitutions and watch which way $h$ goes.
The steps are all 1 apart, so $h$ never gets close to 0 without landing on it. Add the next substitution.
At $h=0$ both ends of the chord are the same point, so there is no chord and the expression is $\tfrac{0}{0}$. After that the steps stay 1 apart and move away from 0 again.
This first value is 1 away from 0, so this chord is well away from the tangent. Add more substitutions and watch which way $h$ goes.
These values of $h$ grow, so each chord is steeper and further from the tangent. They never get near $h=0$.
You can’t just substitute $h=0$.
$$\frac{a^0-1}{0}=\frac{0}{0}$$
That is undefined. The limit as $h\to0$ is the value the expression settles on as $h$ gets closer and closer to 0. So the substitutions must get as close to 0 as you like, but never equal 0.
Look at how far each value is from 0.
(A) $4,\ 2,\ 1,$ $0.5,\ 0.25,\ 0.125$. Halving every time, so it closes in on 0.
(B) $0.05,\ 0.1,\ 0.2,$ $0.4,\ 0.8$. Doubling every time, so it moves away.
(C) lands on 0, then moves away in steps of 1.
(D) $1,\ 2,\ 3,\ 4,\ 5,\ 6$. Moving away.
$\dfrac{a^h-1}{h}$ is the gradient of the chord from $(0,1)$ to $(h,a^h)$ on the graph of $y=a^x$.
$$\frac{a^h-a^0}{h-0}=\frac{a^h-1}{h}$$
As $h\to0$ the chord turns into the tangent at $(0,1)$. Only (A) brings the chord in towards the tangent. In the figure, step through (A) and watch the gradients settle.
To estimate a limit as $h\to0$, substitute values of $h$ that get closer and closer to 0 without equalling 0.
In (A) each value is half the one before, so the values $-4,\ -2,\ -1,$ $-0.5,\ -0.25,$ $-0.125,\ \ldots$ approach 0. (B) and (D) move away from 0, and (C) substitutes $h=0$, which gives $\tfrac{0}{0}$.
How Queensland students answered in 2023
From the QCAA Mathematical Methods subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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