Question 1
The limit of $\dfrac{12^h-1}{h}$ as $h$ approaches 0 is closest to
Not quite. Substituting $h=0$ makes the top $12^0-1=0$, but the bottom is 0 too. $\tfrac{0}{0}$ is undefined, not 0. You need values of $h$ close to 0, not equal to it.
Not quite. 1 is the limit of $\dfrac{e^h-1}{h}$, where the base is $e$. Here the base is 12. You also land near 1 if you use $\log_{10}12\approx1.08$ in place of $\ln12$.
That’s it. Step through the hints below to see why.
Not quite. 3.0 is where you end up if $h$ isn’t small enough. At $h=0.1$, $\dfrac{12^{0.1}-1}{0.1}\approx2.82$, which is nearer 3.0 than 2.5. Closer to 0 the values settle near 2.48.
At $h=0$ both ends of the chord are the same point, so there is no chord, and the expression is $\tfrac{0}{0}$. The limit is about $h$ close to 0, not equal to 0.
Here the chord is still well away from the tangent. Its gradient, , is closest to option (), but $h$ isn’t close enough to 0 to trust it. Move $h$ closer to 0.
Closer. The chord is nearer the tangent, and its gradient, , is nearer $\ln12\approx2.4849$. Keep going towards 0.
On this graph the chord and the tangent look the same, but they aren’t equal: the gradient is , not exactly $\ln12$. From both sides the values settle near 2.48, so the limit is closest to 2.5.
You can’t just substitute $h=0$.
$$\frac{12^0-1}{0}=\frac{0}{0}$$
That is undefined. The limit is the value the expression settles on as $h$ gets closer and closer to 0, from either side.
This is Paper 2, so use your calculator. Substitute small values of $h$ on both sides of 0.
$h=0.01:\ 2.5160$$h=0.001:\ 2.4880$
$h=-0.01:\ 2.4543$$h=-0.001:\ 2.4818$
Don’t stop at $h=0.1$. That gives 2.8209, which is still a long way from the limit.
$\dfrac{12^h-1}{h}$ is the gradient of the chord from $(0,1)$ to $(h,12^h)$ on the graph of $y=12^x$.
$$\frac{12^h-12^0}{h-0}=\frac{12^h-1}{h}$$
As $h\to0$ the chord becomes the tangent at $(0,1)$. That tangent’s gradient is exactly $\ln12\approx2.4849$, which is closest to 2.5.
$\displaystyle \frac{12^{0.001}-1}{0.001}\approx2.4880$$\displaystyle \frac{12^{-0.001}-1}{-0.001}\approx2.4818$
From both sides the values settle near 2.48. The exact limit is $\ln12\approx2.4849$, and the closest option is 2.5.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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