Recognize a derivative inside a limit
A limit with no function named
This limit names no function and no point: the limit of \(\dfrac{(2+h)^4 - 16}{h}\) as \(h\) approaches 0. Compare it with the pattern \(\dfrac{f(a+h) - f(a)}{h}\).
The number that \(h\) is added to is the point \(a\), so \(a = 2\).
The first term of the numerator, \((2+h)^4\), is the output at \(2 + h\). It raises the input to the power 4, so \(f(x) = x^4\).
The number that is subtracted must be \(f(a)\), and \(f(2) = 2^4 = 16\), so the limit is \(f'(2)\) for \(f(x) = x^4\).
Drag \(h\) toward 0. The quotient, the slope of the orange line through \(P = (2, 16)\) and \(Q\), approaches 32, the slope of the tangent at \(P\).
Any limit of this shape
A derivative limit shows three parts. The point \(a\) is the number that \(h\) is added to, and \(f\) is the rule that gives the first term of the numerator. The constant that is subtracted must be \(f(a)\).
If it is, the fraction is the slope of the orange secant line through \(P\) and \(Q\). As \(h\) shrinks, the secant approaches the tangent at \(P\), so the limit is \(f'(a)\).
The denominator must also be the run \(h\). If it is not, the limit is a constant multiple of \(f'(a)\).
Notation
In the \(h\) form, \(a\) is the number that \(h\) is added to.
\[\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\cy{f'(a)}\]In the \(x\) form, \(x\) approaches \(a\), and the denominator is \(x - a\).
\[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\cy{f'(a)}\]If the numerator uses the step \(ch\) but the denominator is \(h\), the fraction is \(c\) times the secant slope, because the run is \(ch\).
\[\lim_{h\to 0}\frac{f(a+ch)-f(a)}{h}=\cy{c\,f'(a)}\]If the denominator is \(ch\) and the numerator uses the step \(h\), the limit is \(\dfrac{f'(a)}{c}\).
When it fails
Change the constant to 15. The numerator \((2+h)^4 - 15\) approaches 1 while the denominator approaches 0, so the fraction grows without bound in size. At \(h = 0.01\) it is about 132, the slope of the red line through \((2, 15)\) and \(Q\).
The limit does not exist, because the subtracted constant is not \(f(a)\), the output at the point.
Example 1: the x form
Identify the function and the point in \(\displaystyle\lim_{x\to 3}\frac{\frac1x-\frac13}{x-3}\).
In the \(x\) form, \(a\) is the number that \(x\) approaches, so \(a = 3\).
\[a=3\]The first term of the numerator, \(\frac1x\), is the output at \(x\), so \(f(x) = \frac1x\).
\[f(x)=\frac1x\]The subtracted constant is \(f(3) = \frac13\), and the denominator is the run \(x - 3\), so the pattern fits.
\[f(3)=\frac13\]The limit is the derivative of \(\frac1x\) at 3.
\[\cy{f'(3)}\]As a check, the tangent at \(P\) has slope \(-\frac19\).
Example 2: a factor in the step
Identify the derivative in \(\displaystyle\lim_{h\to 0}\frac{e^{1+3h}-e}{h}\).
The point is \(a = 1\), the function is \(f(x) = e^x\), and the subtracted constant is \(f(1) = e\).
\[a=1,\quad f(x)=e^x,\quad f(1)=e\]A tempting wrong move: The pattern fits, so the limit is \(f'(1)\).
\[\cr{f'(1)}\]The point \(Q\) is at \(1 + 3h\), so the run of the secant is \(3h\), but the fraction divides the rise by \(h\). For \(h = 0.1\), the orange secant has slope 3.17, but the fraction is 9.51, the slope of the red line.
To put the step \(3h\) in the denominator, multiply and divide by 3.
\[\frac{e^{1+3h}-e}{h}=3\cdot\frac{e^{1+3h}-e}{3h}\]As \(h\) approaches 0, so does \(3h\). The last fraction approaches \(f'(1)\), so the limit is 3 times larger.
\[\cy{3f'(1)}\]
Practice
Identify the derivative in \(\displaystyle\lim_{h\to 0}\frac{\cos(\pi+h)+1}{h}\).
The subtracted constant must be \(f(a)\), the height of \(P\). Adding 1 is the same as subtracting \(-1\).
Solution
The point is \(a = \pi\), because \(h\) is added to \(\pi\).
\[a=\pi\]The function is the cosine, and \(f(\pi) = \cos\pi = -1\). Subtracting \(-1\) is adding 1.
\[\cos\pi=-1\]The limit is the derivative of the cosine at \(\pi\).
\[\cy{f'(\pi)}\]As a check, the tangent at \(P\) is flat, so its slope is 0.
Questions
Questions 1 and 2 use the graph of \(f(x) = \dfrac8x\). Question 3 needs a calculation.
The orange tangent touches the curve at the point \((2, 4)\). What is \(\displaystyle\lim_{h\to 0}\frac{f(2+h)-4}{h}\)?
- −2
- 4
- 2
Answer
−2
What is \(\displaystyle\lim_{h\to 0}\frac{f(2+h)-4}{2h}\)?
- −1
- −2
- −4
Answer
−1
Which derivative is the limit \(\displaystyle\lim_{x\to 3}\frac{x^5-243}{x-3}\)?
- \(f'(3)\) for \(f(x) = x^5\)
- \(f'(243)\) for \(f(x) = x^5\)
- \(f'(5)\) for \(f(x) = x^3\)
Answer
\(f'(3)\) for \(f(x) = x^5\)
Summary
You read the point \(a\) from the number added to \(h\), or from the number that \(x\) approaches.
You read the function from the numerator and check that the subtracted constant is \(f(a)\), because then the fraction is a secant slope.
Together they name the derivative that a limit stands for, times a constant if the denominator is not the step.
This is one of four skills in topic 2.2 of Which & Why: AP Calculus.