When the Two Sides Do Not Meet
A function can approach one value as \(x\) approaches a point from the left and a different value as \(x\) approaches from the right. Its graph then has a jump at that point: nearby function values do not settle toward one common height.
This matters for differentiation because differentiability is a stronger requirement than continuity. If a function has a jump at an interior point, it is not continuous there, so it cannot have a derivative there. The value assigned to the function at the point cannot make two unequal one-sided limits agree.
Definition of a Jump Discontinuity
Let \(f\) be defined at an interior point \(c\) and on some interval around \(c\). The left-hand limit describes the values of \(f(x)\) as \(x\) approaches \(c\) with \(x<c\); the right-hand limit describes them as \(x\) approaches \(c\) with \(x>c\).
The symbols \(L_-\) and \(L_+\) refer to limiting values, not necessarily to \(f(c)\). A function may be assigned a value at \(c\) equal to the left-hand limit, equal to the right-hand limit, or equal to neither. None of those choices removes a jump when \(L_-\ne L_+\).
Continuity at \(c\) requires that \(\lim_{x\to c}f(x)\) exist and equal \(f(c)\). For a two-sided limit to exist, the left-hand and right-hand limits must agree. Thus unequal one-sided limits already rule out continuity, without any calculation involving the assigned point value.
Why a Jump Rules Out Continuity
The reason is the uniqueness of limits: if a two-sided limit exists, approaching from either side must give that same limit. A jump supplies two different one-sided limits instead.
Call it \(L\). By the definition of a two-sided limit, values of \(f(x)\) approach \(L\) as \(x\) approaches \(c\) from either side.
Each one-sided limit must therefore equal \(L\), so \(L_-=L\) and \(L_+=L\).
These equalities imply \(L_-=L_+\), contradicting the defining condition \(L_-\ne L_+\) for a jump. The two-sided limit does not exist, so \(f\) is not continuous at \(c\).
A standard result from differentiability is that if a function has a derivative at an interior point, it is continuous at that point. Therefore a jump discontinuity also rules out a derivative. The function’s value at \(c\) cannot change this conclusion: changing one point does not change either one-sided limit.
Worked Examples
Worked Example: A Step with a Separate Point Value
Suppose $$ f(x)= \begin{cases} 2, & x<1,\\ 4, & x=1,\\ 5, & x>1. \end{cases} $$ Determine whether \(f\) is continuous or differentiable at \(x=1\).
For \(x<1\), the function is constantly \(2\), so $$ \lim_{x\to 1^-}f(x)=2. $$ For \(x>1\), it is constantly \(5\), so $$ \lim_{x\to 1^+}f(x)=5. $$ Both limits are finite, and \(2\ne5\). Thus \(f\) has a jump at \(x=1\). The two-sided limit does not exist, even though \(f(1)=4\) is defined. Therefore \(f\) is not continuous at \(1\), and differentiability is impossible there.
Check: Values immediately to the left remain \(2\), while values immediately to the right remain \(5\). Changing the single value \(f(1)\) to \(2\), \(5\), or any other number would leave these unequal one-sided limits unchanged.
Worked Example: The Point Value Matches One Side
Consider $$ g(x)= \begin{cases} x+2, & x<0,\\ 2, & x=0,\\ 2x+5, & x>0. \end{cases} $$ Test continuity and differentiability at \(x=0\).
From the left, the expression \(x+2\) approaches \(0+2=2\): $$ \lim_{x\to0^-}g(x)=2. $$ From the right, \(2x+5\) approaches \(2(0)+5=5\): $$ \lim_{x\to0^+}g(x)=5. $$ Since \(2\ne5\), the two-sided limit does not exist. The function’s assigned value \(g(0)=2\) agrees with the left-hand limit, but that is not enough for continuity: the right-hand limit is \(5\). Hence \(g\) is not continuous and cannot be differentiable at \(0\).
Check: Substituting \(x=0\) into the left-hand expression gives \(2\), and substituting \(x=0\) into the right-hand expression gives \(5\). These are precisely the calculated one-sided limits, and they are unequal.
Worked Example: Confirming Failure with the Difference Quotient
Let $$ h(x)= \begin{cases} x+2, & x<1,\\ 3, & x=1,\\ 2x+4, & x>1. \end{cases} $$ First use one-sided limits to decide whether \(h'(1)\) can exist. Then check the difference quotient from each side.
The left-hand limit is \(1+2=3\), and the right-hand limit is \(2(1)+4=6\): $$ \lim_{x\to1^-}h(x)=3, \qquad \lim_{x\to1^+}h(x)=6. $$ The limits are unequal, so \(h\) is discontinuous at \(1\) and cannot be differentiable there.
The difference quotient gives a direct check. Write \(x=1+k\), where \(k\ne0\). If \(k<0\), then \(1+k<1\), so $$ h(1+k)=1+k+2=3+k, \qquad \frac{h(1+k)-h(1)}{k} =\frac{(3+k)-3}{k} =\frac{k}{k}=1. $$ If \(k>0\), then \(1+k>1\), so $$ h(1+k)=2(1+k)+4=6+2k, \qquad \frac{h(1+k)-h(1)}{k} =\frac{(6+2k)-3}{k} =\frac{3+2k}{k} =\frac{3}{k}+2. $$ As \(k\) approaches \(0\) from the right, \(3/k+2\) does not approach a finite value. The two-sided derivative limit therefore cannot exist.
Check: The quotient from the left is \(1\), while from the right it is \(3/k+2\). In particular, the right-hand quotient is not approaching the left-hand value. This agrees with the conclusion from the unequal function limits.
A Common Mistake: Checking Only the Point Value or One Side
Suppose a student sees a function with \(f(c)\) equal to the left-hand limit. They may reason that the function is continuous there, then calculate a derivative using only left-hand difference quotients. That checks only one side and misses the jump.
For example, define \(q(x)=1\) for \(x\le0\) and \(q(x)=3\) for \(x>0\). The wrong argument is: “\(q(0)=1\), matching the value on the left, and the left-hand difference quotient is \(0\); therefore \(q'(0)=0\).” The left-hand quotient is indeed $$ \frac{q(h)-q(0)}{h}=\frac{1-1}{h}=0 \qquad (h<0). $$ But from the right, $$ \frac{q(h)-q(0)}{h}=\frac{3-1}{h}=\frac{2}{h} \qquad (h>0), $$ which does not approach a finite number as \(h\to0^+\). Also, the one-sided function limits are \(1\) and \(3\), so \(q\) is not continuous at \(0\). The correction is to compare both one-sided limits before drawing a conclusion about continuity or differentiability.
What to Remember
For a jump discontinuity, the decisive evidence is the pair of one-sided limits. If both exist finitely but are unequal, there is no two-sided limit. That means continuity fails, regardless of the value assigned at the point. Since differentiability at an interior point requires continuity there, no derivative exists at the jump.
Check Your Understanding
For each question, focus on the one-sided limits before considering the point value.
- If \(\lim_{x\to c^-}f(x)=4\) and \(\lim_{x\to c^+}f(x)=4\), do these limits alone establish a jump discontinuity? Explain.
- A function has left-hand limit \(2\), right-hand limit \(7\), and \(f(c)=7\). Is it continuous at \(c\)? Can it be differentiable there?
- Why can changing \(f(c)\) not eliminate a jump when the one-sided limits are unequal?
- A student finds a finite left-hand difference-quotient limit at a jump. Why does that not prove that the derivative exists?
- If the two one-sided limits at an interior point are unequal, what can you conclude about the two-sided limit and continuity?
