Differentiability: corners, cusps and piecewise functions
A graph can be unbroken at a point and still have no slope there. Which shapes cause that, and how do you join pieces of a function so that it has a slope at every join?
Where you are 13 in Derivatives
Why this exists
The derivative at a point defined f'(a) as the limit of the difference quotient and met one place where the limit fails: |x| at 0, a corner. Almost every result ahead begins “suppose f is differentiable”, so we need to know exactly where a derivative exists, what the graph looks like where it does not, and how that connects to Continuity. Piecewise models of paths also need a slope at every join.
Day 1 splits the derivative into a left half and a right half, sorts the ways a continuous graph can lack a derivative into three shapes, and proves that a derivative forces continuity. Day 2 turns this into a test for piecewise functions and uses it to choose parameters so that pieces join smoothly.
Definition
One-sided derivatives. The left-hand derivative and right-hand derivative of f at a are
each provided the limit exists as a finite number. The left-hand derivative is the limit of the slopes of secant lines from P(a, f(a)) to points of the graph on its left; the right-hand derivative uses points on its right.
Differentiable. A function is differentiable at a when f'(a) exists (Definition of the derivative). It is differentiable on an open interval when it is differentiable at every point of it, and differentiable on a closed interval [a, b] when it is differentiable on (a, b) and f'_+(a) and f'_-(b) exist, the same pattern as Continuity on an interval.
Three shapes. Suppose f is continuous at a but not differentiable there, and that on each side the difference quotient has a limit, finite or infinite. The graph at (a, f(a)) has
- a corner when the two one-sided limits of the quotient are different and at least one of them is finite;
- a cusp when the quotient approaches \infty on one side and -\infty on the other;
- a vertical tangent when the quotient approaches \infty on both sides, or -\infty on both sides. The tangent line is then the vertical line x = a, which has no slope.
The remaining way to lose a derivative is to lose continuity: a hole, a jump or an infinite break (Discontinuities and how to remove them).
Intuition
Zoom in on a graph where it is differentiable and it straightens into a line that is not vertical. Each failure refuses to straighten (Figure 1). A corner stays two rays with different slopes at every zoom. At a cusp both branches swing up against the vertical line through the point and meet in a sharp spike. At a vertical tangent the graph also swings toward the vertical line, but passes straight through, still rising (or still falling).
A break is worse. Across a jump at a, a secant from P keeps a rise near the size of the jump while its run shrinks to 0, so its slope grows without bound. That is why a derivative forces continuity.
Results
Day 1: one-sided derivatives and continuity
One-sided test. f'(a) exists exactly when f'_-(a) and f'_+(a) both exist and are equal, and then f'(a) is their common value.
Differentiable implies continuous. If f'(a) exists, then f is continuous at a. One side at a time: if f'_-(a) exists, then \lim_{x \to a^-} f(x) = f(a), and if f'_+(a) exists, then \lim_{x \to a^+} f(x) = f(a). Equivalently, a function that is not continuous at a is not differentiable at a.
The converse is false. A function can be continuous at a and not differentiable there: every corner, cusp and vertical tangent is an example.
Day 2: piecewise functions
Piecewise test. Suppose that near a, f(x) = g(x) for x < a and f(x) = k(x) for x > a, where g and k are formulas differentiable on an open interval containing a, and f(a) is the value of one of them. Then f is differentiable at a exactly when
and then f'(a) = g'(a) = k'(a). The first condition is continuity at a; the second is matching slopes.
Two conditions per join. Continuity gives one equation and matching slopes a second, so two unknown parameters at a join are usually fixed exactly.
Why it is true
One-sided test. A two-sided limit exists exactly when both one-sided limits exist and agree (One sided limits and when a limit exists). Apply that to the difference quotient.
Differentiable implies continuous. The derivative at a point gave this in one line; here it is with the one-sided form. Suppose f'_+(a) exists. For h > 0,
As h \to 0^+ the first factor approaches 0 and the second approaches the finite number f'_+(a), so by the product law for limits the right side approaches 0 \cdot f'_+(a) = 0. Hence \lim_{h \to 0^+} f(a + h) = f(a), which is continuity from the right. The left side is the same with h < 0, and continuity from both sides is continuity. Finiteness matters: an unbounded quotient defeats the product law, as beside a jump (Example 4).
The converse is false. Example 1 below is continuous at 1 with no derivative there.
Piecewise test. First suppose g(a) = f(a). For h < 0 the point a + h is on the left, so f(a + h) = g(a + h) and
which approaches g'(a) as h \to 0^-. So f'_-(a) = g'(a), and likewise k(a) = f(a) gives f'_+(a) = k'(a). If both values equal f(a), the one-sided test says f is differentiable exactly when g'(a) = k'(a). If instead g(a) or k(a) differs from f(a), then since g and k are continuous at a (being differentiable), the one-sided limit of f on that side is not f(a), so f is not continuous, hence not differentiable, at a. The slopes of the formulas come from the limit itself: Finding a derivative from the definition found 2x, 3x^2 and \displaystyle \frac{1}{2\sqrt{x}}, and \displaystyle \frac{1}{x} gives \displaystyle -\frac{1}{x^2} the same way. By the limit laws, a constant multiple or sum of formulas has that multiple or sum of slopes (proved as rules in The power rule and basic differentiation rules).
Worked examples
Day 1: a corner where both sides rise
Show that f(x) = x^2 + |x - 1| is continuous at x = 1 but not differentiable there, and find both one-sided derivatives.
SolutionHere f(1) = 1, and f is continuous at 1 because it is a sum of a polynomial and the absolute value of a polynomial, both continuous everywhere. For h \ne 0,
For h > 0, \displaystyle \frac{|h|}{h} = 1 and the quotient is 3 + h, which approaches 3. For h < 0, \displaystyle \frac{|h|}{h} = -1 and the quotient is 1 + h, which approaches 1. So
They differ, so f'(1) does not exist: a corner at (1, 1). Unlike |x|, the graph rises on both sides, turning sharply from slope 1 to slope 3, with half-tangents y = x on the left and y = 3x - 2 on the right (Figure 1, left).
Day 1: a cusp and a vertical tangent
Each function is continuous at the given point. Classify the point. (a) \displaystyle f(x) = \sqrt{|x - 4|} at x = 4 (b) \displaystyle g(x) = \sqrt[3]{x + 1} at x = -1
Solution(a) f(4) = 0, so the quotient is \displaystyle \dfrac{\sqrt{|h|}}{h}. For h > 0 it is \displaystyle \dfrac{\sqrt{h}}{h} = \dfrac{1}{\sqrt{h}}, which grows without bound. For h < 0, |h| = -h and the quotient is \displaystyle \dfrac{\sqrt{-h}}{h} = -\dfrac{1}{\sqrt{-h}}, which decreases without bound. So
and neither one-sided derivative exists as a number. Opposite infinities make a cusp: the graph comes down steeply to (4, 0) and leaves just as steeply upward (Figure 1, middle). The same shape returns with x^{2/3} in The power rule and basic differentiation rules.
(b) g(-1) = 0 and \displaystyle g(-1 + h) = \sqrt[3]{h}, so the quotient is
for h \ne 0. Since h^2 > 0, this is positive on both sides and grows without bound as h \to 0 from either side. Both one-sided quotients approach \infty, so the graph has a vertical tangent, the line x = -1 (Figure 1, right). It has a tangent line, but a vertical line has no slope, so g'(-1) does not exist.
The signs tell the shapes apart: in (a) the left secants slope down and the right ones up, so the graph reverses direction; in (b) all slope up and the graph keeps rising.
Day 1: reading differentiability from a graph
Figure 2 shows the graph of
At each of x = -2, 1, 5 and 6, decide whether p is continuous and whether it is differentiable, and name the shape.
SolutionAt x = -2: p(-2) = 3 from the first piece, and the second piece approaches \displaystyle \sqrt{9} = 3, so p is continuous. For h < 0 the quotient is \displaystyle \frac{(-2 + h) + 5 - 3}{h} = 1, so p'_-(-2) = 1. For h > 0, 3(1 - (-2 + h)) = 9 - 3h, and multiplying by the conjugate gives
which approaches \displaystyle -\frac{3}{6} = -\frac{1}{2}. The one-sided derivatives 1 and \displaystyle -\frac{1}{2} differ: a corner.
At x = 1: p(1) = 0 and both neighboring pieces approach 0, so p is continuous. For h < 0 the quotient is \displaystyle \frac{\sqrt{-3h}}{h} = -\frac{\sqrt{3}}{\sqrt{-h}}, which decreases without bound; for h > 0 it is \displaystyle \frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}, which grows without bound. Opposite infinities: a cusp.
At x = 5: p(5) = 2, but as x \to 5^+, \displaystyle p(x) \to 4 + \sqrt[3]{-1} = 3. A jump: p is not continuous at 5, so it is not differentiable there.
At x = 6: p(6) = 4 and the quotient is \displaystyle \frac{\sqrt[3]{h}}{h} = \frac{1}{\sqrt[3]{h^2}}, which approaches \infty from both sides: a vertical tangent, x = 6. Everywhere else in (-4, 7) the formula in use is differentiable, so these four points are the only interior points where p' fails to exist.
Day 2: matching slopes is not enough
Let f(x) = \begin{cases} x^2, & x < 2 \\ 4x - 3, & x \ge 2 \end{cases}. Ruwan differentiates each piece, gets 2x and 4, notes that both equal 4 at x = 2, and concludes f'(2) = 4. Is he right?
SolutionCheck continuity first. f(2) = 4(2) - 3 = 5, but \lim_{x \to 2^-} x^2 = 4. The graph jumps, so f is not continuous at 2 and not differentiable there. Ruwan is wrong.
The quotients show what goes wrong. On the right, \displaystyle \frac{4(2 + h) - 3 - 5}{h} = 4, so f'_+(2) = 4. On the left,
As h \to 0^- the numerator approaches -1 and the denominator approaches 0 through negative numbers, so the quotient grows without bound: the left secants run from (2, 5) down to the parabola near (2, 4) and are nearly vertical. The slope of the formula x^2 is not the left-hand derivative of f, because the piecewise test needs g(2) = f(2), and 4 \ne 5.
Changing the second piece to 4x - 4 repairs it: value 4 from both sides, slopes 4 and 4, derivative 4. The line y = 4x - 4 is the tangent line to y = x^2 at (2, 4); a line that joins a curve smoothly is tangent to it.
Day 2: a function with two joins
Find where f(x) = \begin{cases} 3x + 3, & x < -1 \\ x^2 + x, & -1 \le x \le 2 \\ 5x - 4, & x > 2 \end{cases} is differentiable, and write f'(x).
SolutionEach piece is a polynomial, so f is differentiable at every point other than the joins. At x = -1: f(-1) = 1 - 1 = 0 and the left piece approaches 3(-1) + 3 = 0, so f is continuous. With continuity settled, the piecewise test applies: the slopes are 3 from the left and 2(-1) + 1 = -1 from the right. They differ, so f'(-1) does not exist; the graph has a corner.
At x = 2: f(2) = 4 + 2 = 6 and the right piece approaches 5(2) - 4 = 6, so f is continuous. The slopes are 2(2) + 1 = 5 and 5. They match, so f'(2) = 5. Hence f is differentiable at every x except -1, and
The derivative is continuous at 2, where the join is smooth, and jumps at -1: a corner of f shows up as a jump in f'.
Day 2: choosing two parameters
Find a and b so that \displaystyle f(x) = \begin{cases} \sqrt{x}, & 0 \le x \le 4 \\ ax^2 + bx, & x > 4 \end{cases} is differentiable on (0, \infty).
SolutionAway from x = 4 each piece is differentiable on (0, \infty), so only the join matters. Continuity at 4 needs the right piece to reach \displaystyle \sqrt{4} = 2:
For the slopes, \displaystyle \frac{d}{dx}\big[\sqrt{x}\,\big] = \frac{1}{2\sqrt{x}}, which is \displaystyle \frac{1}{4} at x = 4, and \displaystyle \frac{d}{dx}\big[ax^2 + bx\big] = 2ax + b, which is 8a + b. Matching them:
From the second equation \displaystyle b = \frac{1}{4} - 8a. Substituting, 16a + 1 - 32a = 2, so \displaystyle a = -\frac{1}{16} and \displaystyle b = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}. Check: \displaystyle 16\left(-\frac{1}{16}\right) + 4 \cdot \frac{3}{4} = 2 and \displaystyle 8\left(-\frac{1}{16}\right) + \frac{3}{4} = \frac{1}{4}, so \displaystyle f'(4) = \frac{1}{4}. Continuity alone has infinitely many solutions, each with a corner at 4; the slope equation picks the smooth one.
Day 2: Ruwan’s greenway
Ruwan maps a greenway, a paved path for walking and biking, from above, with x and y in hundreds of feet. The path runs along y = 1 for x \le 0 and along y = x for x \ge 4. Between them he plans the curve y = 1 + ax^3 + bx^2 for 0 < x < 4. For the path to have no sharp turn, the whole path must be the graph of a differentiable function. Find a and b.
SolutionWrite c(x) = 1 + ax^3 + bx^2, so c'(x) = 3ax^2 + 2bx. At the join x = 0, c(0) = 1 matches the height of y = 1, and c'(0) = 0 matches its slope. Both hold for every a and b; the form of the curve was chosen for that. So both conditions come from the join at x = 4. The height must be 4 and the slope must be 1, the slope of y = x:
Doubling the second equation gives 96a + 16b = 2, and subtracting the first gives 32a = -1, so \displaystyle a = -\frac{1}{32}. Then 16b = 3 - 64a = 5, so \displaystyle b = \frac{5}{16}. The curve is
Check: c(4) = 1 - 2 + 5 = 4 and \displaystyle c'(4) = -\frac{3}{2} + \frac{5}{2} = 1. Also \displaystyle c'(x) = \frac{x(20 - 3x)}{32} is positive on (0, 4), so the curve climbs steadily (Figure 3).
A continuous join without a derivative would be a kink, a sudden change of heading; matching slopes removes it.
Boundaries
- The three shapes do not cover every failure. The difference quotient may have no limit at all, finite or infinite. For \displaystyle f(x) = x \sin \frac{1}{x} with f(0) = 0, the function is continuous at 0 (squeeze it between -|x| and |x|), yet the quotient \displaystyle \sin \frac{1}{h} swings between -1 and 1 forever as h \to 0.
- A vertical tangent is a tangent line with no derivative. “The graph has a tangent line at a” and “f is differentiable at a” differ in exactly this case.
- The piecewise shortcut needs continuity first, as Example 4 shows, and it needs each formula to be differentiable at the join. If a formula is not (such as \displaystyle \sqrt{x} at 0), compute the one-sided quotient directly.
- At an endpoint of the domain only one side exists, and even that one-sided derivative may fail: for \displaystyle \sqrt{x} at 0 the right quotient \displaystyle \frac{1}{\sqrt{h}} grows without bound.
- Continuity is much weaker than differentiability. Some functions are continuous at every real number and differentiable at none (stated without proof).
Where this goes next
Points where f' does not exist reappear in Maxima, minima and critical points: a maximum or minimum can sit at a corner or a cusp, as the lowest point of \displaystyle \sqrt{|x - 4|} does. The derivative as a function reads corners of f as jumps in f', and Implicit differentiation meets vertical tangents on curves such as circles. The continuity step with parameters is treated on its own in Making a piecewise function continuous, and the functions themselves in Piecewise and absolute value functions.
Exercises
Day 1: one-sided derivatives and the three shapes
- Find f'_-(2) and f'_+(2) for f(x) = |x^2 - 2x|. Is f differentiable at 2?
- Find both one-sided derivatives of f(x) = 3x + |x + 2| at x = -2.
- Show that f(x) = |x|^3 is differentiable at 0 and find f'(0).
- Classify the point x = -2 on the graph of \displaystyle f(x) = \sqrt[3]{(x + 2)^2} as a corner, a cusp or a vertical tangent.
- Classify the point x = 1 on the graph of \displaystyle f(x) = \sqrt[5]{x - 1} and give the tangent line there.
- Classify the point x = -1 on the graph of \displaystyle f(x) = 2 - \sqrt{|x + 1|}.
- Find f'_+(0) for \displaystyle f(x) = x\sqrt{x} on [0, \infty).
Day 1: Figure 2 and a step function
- For the function p of Example 3, find p'_+(-4), p'_-(5) and p'_-(7).
- The formula \displaystyle 4 + \sqrt[3]{x - 6} has a slope at x = 5. Explain why p still has no right-hand derivative at 5 by finding the limit of the right-hand quotient.
- For q(x) = \lfloor x \rfloor, find both one-sided limits of the difference quotient at x = 3, and explain how they fit the theorem that a one-sided derivative forces one-sided continuity.
Day 2: testing piecewise functions
Decide whether each function is differentiable at the join, and find the derivative there if it exists.
- f(x) = \begin{cases} x^2 + 1, & x \le 1 \\ 2x, & x > 1 \end{cases}
- f(x) = \begin{cases} x^3, & x < 1 \\ 3x - 1, & x \ge 1 \end{cases}
- f(x) = \begin{cases} 4 - x^2, & x \le 1 \\ \dfrac{2}{x} + 1, & x > 1 \end{cases}
- f(x) = \begin{cases} x^2 - 4x, & x < 3 \\ 3 - 2x, & x \ge 3 \end{cases}
Day 2: choosing parameters
Find the constants that make each function differentiable everywhere.
- f(x) = \begin{cases} ax^2 + b, & x \le 1 \\ \dfrac{1}{x}, & x > 1 \end{cases}
- f(x) = \begin{cases} x^2 + ax, & x \le 1 \\ bx^3, & x > 1 \end{cases}
- f(x) = \begin{cases} ax + b, & x \le -1 \\ x^3 - 2x, & x > -1 \end{cases}
- Chathura plans a second path that runs along y = 0 for x \le 0 and along y = x - 2 for x \ge 6, joined by y = ax^3 + bx^2 for 0 < x < 6. Find a and b so the path has no sharp turn.
- Show that no value of a makes f(x) = \begin{cases} ax^2, & x \le 1 \\ 2x + 1, & x > 1 \end{cases} differentiable at 1.
- Nadeesha joins y = x^2 for x \le c to y = 6x - 9 for x > c. Find c so that the resulting function is differentiable everywhere.
Reasoning
- Chathura says that if f'_-(a) and f'_+(a) both exist, then f is continuous at a, even when the two numbers differ. Is he right? Explain.
- Give a function that is continuous at 2 with a corner there, and one with a cusp there. Justify each.
- Suppose g is differentiable at a and g(a) = 0. Show that |g| has one-sided derivatives -|g'(a)| and |g'(a)| at a. When is |g| differentiable at a?
- Nadeesha says that at a jump discontinuity the difference quotient must grow without bound on at least one side. Prove it.
Answers
- f'_-(2) = -2, f'_+(2) = 2; not differentiable (a corner). f(2) = 0 and (2 + h)^2 - 2(2 + h) = h(2 + h), so the quotient is \displaystyle \frac{(2 + h)|h|}{h} for small h, which is 2 + h for h > 0 and -(2 + h) for h < 0.
- f'_-(-2) = 2 and f'_+(-2) = 4. f(-2) = -6 and f(-2 + h) = -6 + 3h + |h|, so the quotient is \displaystyle 3 + \frac{|h|}{h}.
- f'(0) = 0. The quotient is \displaystyle \frac{|h|^3}{h} = \frac{h^2|h|}{h} = h|h|, which approaches 0 from both sides.
- A cusp. f(-2) = 0 and the quotient is \displaystyle \frac{\sqrt[3]{h^2}}{h} = \frac{1}{\sqrt[3]{h}}, which approaches \infty as h \to 0^+ and -\infty as h \to 0^-. The function is continuous at -2.
- A vertical tangent, the line x = 1. The quotient is \displaystyle \frac{\sqrt[5]{h}}{h} = \frac{1}{\sqrt[5]{h^4}}, which is positive for h \ne 0 and approaches \infty from both sides.
- A cusp, a sharp peak at (-1, 2). The quotient is \displaystyle -\frac{\sqrt{|h|}}{h}: for h > 0 it is \displaystyle -\frac{1}{\sqrt{h}} \to -\infty, and for h < 0 it is \displaystyle \frac{1}{\sqrt{-h}} \to \infty.
- f'_+(0) = 0. The quotient is \displaystyle \frac{h\sqrt{h}}{h} = \sqrt{h} for h > 0, which approaches 0.
- p'_+(-4) = 1, \displaystyle p'_-(5) = \frac{1}{4}, \displaystyle p'_-(7) = \frac{1}{3}. On the line the quotient is 1. At 5 the value p(5) = 2 comes from \displaystyle \sqrt{x - 1}, so the left quotient is \displaystyle \frac{\sqrt{4 + h} - 2}{h} = \frac{1}{\sqrt{4 + h} + 2} \to \frac{1}{4}. At 7 the quotient is \displaystyle \frac{\sqrt[3]{1 + h} - 1}{h}; with \displaystyle u = \sqrt[3]{1 + h}, h = u^3 - 1 = (u - 1)(u^2 + u + 1), so the quotient is \displaystyle \frac{1}{u^2 + u + 1} \to \frac{1}{3}.
- The right-hand quotient approaches \infty. For small h > 0, p(5 + h) \to 3 while p(5) = 2, so the numerator approaches 1 and the denominator 0^+. The quotient uses p(5) = 2, from the left piece, so the right formula’s slope is beside the point.
- Right limit 0; the left quotient grows without bound. For 0 < h < 1 the quotient is \displaystyle \frac{3 - 3}{h} = 0, so q'_+(3) = 0, and q is continuous from the right at 3. For -1 < h < 0 it is \displaystyle \frac{2 - 3}{h}, so \displaystyle -\frac{1}{h} \to \infty; q is not continuous from the left, so it has no left-hand derivative.
- Differentiable, f'(1) = 2. f(1) = 2 and the right piece approaches 2, so f is continuous; the slopes 2x at 1 and 2 are both 2.
- Not differentiable at 1. f(1) = 2 but \lim_{x \to 1^-} x^3 = 1, a jump. The slopes 3 and 3 match, but the left quotient \displaystyle \frac{(1 + h)^3 - 2}{h} approaches \infty as h \to 0^-.
- Differentiable, f'(1) = -2. Both pieces give 3 at 1, and the slopes are -2x \to -2 and \displaystyle -\frac{2}{x^2} \to -2.
- Not differentiable at 3: a corner. f(3) = -3 and the left piece approaches 9 - 12 = -3, so f is continuous; the slopes are 2x - 4 = 2 and -2.
- \displaystyle a = -\frac{1}{2}, \displaystyle b = \frac{3}{2}. Continuity: a + b = 1. Slopes: \displaystyle 2a = -\frac{1}{1^2} = -1.
- \displaystyle a = -\frac{1}{2}, \displaystyle b = \frac{1}{2}. Continuity: 1 + a = b. Slopes: 2 + a = 3b = 3 + 3a, so \displaystyle a = -\frac{1}{2}.
- a = 1, b = 2. The right piece gives -1 + 2 = 1 and slope 3 - 2 = 1 at -1, so a = 1 and -a + b = 1. The line y = x + 2 is tangent to y = x^3 - 2x at (-1, 1).
- \displaystyle a = -\frac{1}{108}, \displaystyle b = \frac{1}{6}. The join at 0 is smooth for every a, b. At 6: height 216a + 36b = 4 and slope 108a + 12b = 1. Tripling the second and subtracting the first gives 108a = -1; then 36b = 4 + 2 = 6.
- Continuity needs a = 3, and matching slopes needs 2a = 2, so a = 1. No a does both. With a = 3 the function is continuous with a corner (slopes 6 and 2).
- c = 3. Continuity: c^2 = 6c - 9, so (c - 3)^2 = 0. Slopes: 2c = 6. Both give c = 3, so the line is tangent to the parabola at (3, 9).
- Yes. A finite f'_-(a) gives continuity from the left and a finite f'_+(a) continuity from the right (Why it is true); together they are continuity. Example 1 is continuous at 1 but, with 1 \ne 3, not differentiable.
- For example |x - 2|, with quotient \displaystyle \frac{|h|}{h} and one-sided limits -1 and 1 (a corner), and \displaystyle \sqrt{|x - 2|}, with quotient \displaystyle \frac{\sqrt{|h|}}{h} approaching \infty on the right and -\infty on the left (a cusp). Both are continuous, being built from continuous functions.
- Since g(a) = 0, the quotient is \displaystyle \left|\frac{g(a + h) - g(a)}{h}\right| \cdot \frac{|h|}{h}. The first factor approaches |g'(a)|; the second is 1 for h > 0 and -1 for h < 0. So the one-sided derivatives are -|g'(a)| and |g'(a)|. They agree only when g'(a) = 0, and then |g| is differentiable at a with derivative 0; otherwise it has a corner.
- At a jump the left and right limits L and R differ, so at least one, say L, differs from f(a). As h \to 0^- the numerator approaches the nonzero number L - f(a) while the denominator approaches 0, so the quotient grows without bound in size, toward \infty or -\infty. The right side is the same when R \ne f(a).