Check whether the defining limit exists
A peak with two slopes
The graph is \(f(x) = 3 - |x - 2|\), with a peak at \(P = (2, 3)\). To test \(f'(2)\), find the slope of the orange secant line from \(P\) to a point \(Q\) at \(2 + h\), which is the difference quotient \(\dfrac{f(2+h) - 3}{h}\).
Choose \(h = 0.5\). The point \(Q\) is to the right of the peak, at height 2.5, so the rise is \(-0.5\) and the slope is \(-0.5 \div 0.5 = -1\). The vertical leg is red because the rise is negative.
Choose \(h = -0.5\). The point \(Q\) is to the left of the peak, also at height 2.5. The rise is again \(-0.5\), but the step is \(-0.5\), so the slope is \(+1\).
Shrink h on each side. The slope from the right stays −1 and the slope from the left stays 1, so the two sides approach different numbers.
Two sides of a point
A derivative at \(a\) is a two-sided limit, so \(h\) can approach 0 from either side. The orange secant line uses \(h > 0\), with \(Q\) to the right of \(P\), and the faint one uses \(h < 0\), with its point to the left.
Press Shrink both steps. The two secant lines turn toward the red half-lines, whose slopes are the limits from the right and from the left.
If both limits exist and are equal, \(f'(a)\) exists and has that value. Here the half-lines are red because their slopes differ, so the graph has a corner at \(a\) and \(f'(a)\) does not exist.
Notation
The right-hand derivative uses steps \(h > 0\).
\[f'_{+}(a)=\lim_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}\]The left-hand derivative uses steps \(h < 0\).
\[f'_{-}(a)=\lim_{h\to 0^{-}}\frac{f(a+h)-f(a)}{h}\]The derivative exists only when both one-sided limits are finite numbers and are equal, and then it is their common value.
\[\cy{f'(a)}=f'_{-}(a)=f'_{+}(a)\]If the two limits differ, or if either one does not exist, write that \(f'(a)\) does not exist.
If a formula changes at \(a\), use the formula for each side in its own quotient. The value \(f(a)\) comes from the formula that includes \(a\).
When it fails
Replace the corner with the cube root \(f(x) = 1 + \sqrt[3]{x-2}\) at \(a = 2\). The graph has no corner, but its tangent at \(a\) is the vertical orange line.
The quotient is \(\dfrac{\sqrt[3]{h}}{h} = \dfrac{1}{h^{2/3}}\), which is 100 for \(h = 0.001\) and for \(h = -0.001\), and 10,000 for \(h = 0.000001\). It grows without bound from both sides, so the limit does not exist.
Example 1: two formulas that join smoothly
Let \(f(x) = x^2\) for \(x \le 1\) and \(f(x) = 2x - 1\) for \(x > 1\). Does \(f'(1)\) exist?
The first formula covers \(x \le 1\), so it gives \(f(1) = 1^2 = 1\). The second formula also gives \(2\cdot 1 - 1 = 1\), so the two pieces meet.
\[f(1)=1\]For \(h < 0\), use \(x^2\). Because \((1+h)^2 - 1 = 2h + h^2\), the quotient is \(2 + h\), so its limit from the left is 2.
\[\frac{(1+h)^2-1}{h}=2+h \;\to\; 2\]For \(h > 0\), use \(2x - 1\). The quotient is 2 for every \(h\), so its limit from the right is 2.
\[\frac{2(1+h)-1-1}{h}=2 \;\to\; 2\]Both one-sided limits are 2, so the derivative exists.
\[\cy{f'(1)=2}\]
Example 2: two formulas that meet at a corner
Let \(g(x) = x^3\) for \(x \le 1\) and \(g(x) = 5x - 4\) for \(x > 1\). Does \(g'(1)\) exist?
The first formula covers \(x \le 1\), so it gives \(g(1) = 1^3 = 1\). The second formula also gives \(5\cdot 1 - 4 = 1\), so the two pieces meet.
\[g(1)=1\]A tempting wrong move: Both formulas give the same value at 1, so the graph is connected and \(g'(1)\) exists.
\[\cr{g'(1)\text{ exists}\ ?}\]The pieces meet at \((1, 1)\), but the red half-lines show slope 3 on the left and slope 5 on the right. Two different slopes meet there, so the graph has a corner.
For \(h < 0\), use \(x^3\), where \((1+h)^3 - 1 = 3h + 3h^2 + h^3\). For \(h > 0\), use \(5x - 4\).
\[\begin{aligned}\frac{(1+h)^3-1}{h}&=3+3h+h^2\\ &\to 3\\ \frac{5(1+h)-4-1}{h}&=5\\ &\to 5\end{aligned}\]The one-sided limits are different, so the derivative does not exist.
\[\cr{3\neq 5:\ g'(1)\text{ does not exist}}\]
Practice
Does \(k'(0)\) exist for \(k(x) = x\,|x|\)? If it does, find it.
Write \(|h|\) as \(h\) when \(h > 0\) and as \(-h\) when \(h < 0\). Then find the quotient from each side.
Solution
Because \(k(0) = 0\) and \(k(h) = h|h|\), the quotient is \(h|h|\) divided by \(h\), which is \(|h|\).
\[\frac{k(h)-k(0)}{h}=|h|\]From the right, \(|h|\) equals \(h\), which approaches 0. From the left, \(|h|\) equals \(-h\), which also approaches 0.
\[k'_{+}(0)=0, \qquad k'_{-}(0)=0\]The one-sided limits are equal, so the derivative exists and the tangent at 0 is flat.
\[\cy{k'(0)=0}\]
Questions
Questions 1 and 2 use the graph of \(f(x) = |x^2 - 4|\), which meets the x-axis at \(x = 2\). Question 3 needs a calculation.
What is the limit of \(\dfrac{f(2+h)-f(2)}{h}\) as \(h\) approaches 0 from the left?
- −4
- 4
- 0
Answer
−4
Does \(f'(2)\) exist?
- No, because the one-sided slopes are −4 and 4
- Yes, because \(f\) is continuous at \(x = 2\)
- Yes, with value 0, the average of −4 and 4
- Yes, because both one-sided limits exist
Answer
No, because the one-sided slopes are −4 and 4
Let \(h(x) = x^2\) for \(x \le 3\) and \(h(x) = 6x - 9\) for \(x > 3\). Does \(h'(3)\) exist?
- Yes, and \(h'(3) = 6\)
- No, because the two pieces have different formulas
- Yes, and \(h'(3) = 9\)
Answer
Yes, and \(h'(3) = 6\)
Summary
You compute the quotient for \(h > 0\) and for \(h < 0\), with the formula that applies on each side.
The derivative exists only when both one-sided limits are finite and equal, because a single tangent needs one slope.
You use this test at a corner or where two formulas join, and when the limits differ, the derivative does not exist.
This is one of four skills in topic 2.2 of Which & Why: AP Calculus.