When the Two Sides Do Not Meet
A piecewise function can approach one value as its input comes toward a breakpoint from the left, and a different value as the input comes from the right. When both values are finite, the graph has a finite jump at that point. The function may still have a value at the breakpoint, but that single value cannot make the two sides meet.
This tutorial focuses on how to calculate and interpret that mismatch. The key is to keep three quantities separate: the left-hand limit, the right-hand limit, and the function’s value at the breakpoint.
One-Sided Limits and a Finite Jump
Let \(c\) be an interior point of a function’s domain, so the function is defined at inputs on both sides of \(c\) sufficiently close to \(c\). The left-hand limit of \(f(x)\) at \(c\) is \(L\) if \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x<c\) sufficiently close to \(c\). The right-hand limit is defined in the same way, using \(x>c\).
The symbols \(c^-\) and \(c^+\) indicate that the input approaches \(c\) from the left and right, respectively. To compute each limit for a piecewise-defined function, use the formula that applies on that side. The value \(f(c)\) is not used to calculate either one-sided limit.
At an interior point, continuity requires the left- and right-hand limits to agree and to equal \(f(c)\). Thus unequal one-sided limits already rule out continuity, whatever value the function has at \(c\). This is different from a removable discontinuity, where the two sides approach the same finite number but the function’s value may be missing or different.
Worked Example: A Jump Between Two Simple Formulas
Worked Example: A Jump Between Two Simple Formulas
Define $$ f(x)= \begin{cases} x+2, & x<1,\\ 4, & x=1,\\ x^2, & x>1. \end{cases} $$ Find both one-sided limits at \(x=1\), classify the discontinuity, and state whether changing \(f(1)\) could make \(f\) continuous there.
Left-hand limit. For \(x<1\), the applicable formula is \(x+2\). Therefore, $$ \lim_{x\to 1^-}f(x) =\lim_{x\to 1^-}(x+2) =1+2 =3. $$
Right-hand limit. For \(x>1\), use \(x^2\): $$ \lim_{x\to 1^+}f(x) =\lim_{x\to 1^+}x^2 =1^2 =1. $$
Both limits are finite, but \(3\ne 1\). Hence \(f\) has a finite jump discontinuity at \(x=1\). The signed jump is \(1-3=-2\), and its magnitude is \(2\). The assigned value is \(f(1)=4\), which is neither one-sided limit; changing it would still leave the unequal limits \(3\) and \(1\).
Check: On the left, the difference from the proposed limit is \(|(x+2)-3|=|x-1|\), which tends to \(0\). On the right, \(|x^2-1|=|x-1||x+1|\), which tends to \(0\) as \(x\to1^+\). These confirm the two computed limits.
Why Changing the Point Value Cannot Repair a Jump
The reason is built into the meaning of a limit. A limit describes values of \(f(x)\) for inputs close to \(c\), but different from \(c\). Changing just \(f(c)\) does not change any of those nearby values. In particular, it cannot alter either one-sided limit.
Justification. Continuity at \(c\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that $$ |f(x)-f(c)|<\varepsilon \quad\text{whenever}\quad 0<|x-c|<\delta. $$ For inputs satisfying \(c-\delta<x<c\), the same inequality holds, so the values of \(f(x)\) approach \(f(c)\) from the left. Thus the left-hand limit is \(f(c)\). For inputs satisfying \(c<x<c+\delta\), it also holds, so the right-hand limit is \(f(c)\). Therefore both one-sided limits must equal one another and \(f(c)\). Unequal one-sided limits contradict this necessary condition for continuity.
This argument also explains why choosing \(f(c)\) to match one side is not enough. If \(f(c)=L_-\), the right-hand limit remains \(L_+\ne L_-\); if \(f(c)=L_+\), the left-hand limit remains \(L_-\ne L_+\). Any other assigned value matches neither side.
Worked Example: A Jump in a Rational Piece
Worked Example: A Jump in a Rational Piece
Consider $$ g(x)= \begin{cases} \dfrac{x^2-4}{x-2}, & x<2,\\ 0, & x=2,\\ 3x-1, & x>2. \end{cases} $$ Calculate the one-sided limits at \(2\) and decide whether any replacement for \(g(2)\) makes the function continuous.
For \(x<2\), the denominator is nonzero, and factoring gives $$ \frac{x^2-4}{x-2} =\frac{(x-2)(x+2)}{x-2} =x+2. $$ Thus $$ \lim_{x\to2^-}g(x) =\lim_{x\to2^-}(x+2) =2+2 =4. $$
For \(x>2\), the formula is \(3x-1\), so $$ \lim_{x\to2^+}g(x) =\lim_{x\to2^+}(3x-1) =3(2)-1 =6-1 =5. $$ The limits are finite and unequal: \(4\ne5\). The discontinuity is a finite jump, with signed jump \(5-4=1\) and magnitude \(1\). Replacing \(g(2)=0\) by any other number leaves these one-sided limits unchanged, so no replacement can make \(g\) continuous.
Check: For \(x<2\), cancellation gives the exact expression \(x+2\), whose difference from \(4\) is \(x-2\), tending to \(0\). For \(x>2\), the difference from \(5\) is \((3x-1)-5=3(x-2)\), also tending to \(0\). The side-specific calculations agree with the stated limits.
Worked Example: The Breakpoint Value Is Not a Side Limit
Worked Example: The Breakpoint Value Is Not a Side Limit
Let $$ h(x)= \begin{cases} -2, & x<0,\\ 7, & x=0,\\ 3, & x>0. \end{cases} $$ Find the jump at \(0\), and explain what happens if the value \(h(0)\) is changed to \(-2\).
Every value on the left of \(0\) is \(-2\), so $$ \lim_{x\to0^-}h(x)=-2. $$ Every value on the right is \(3\), so $$ \lim_{x\to0^+}h(x)=3. $$ The limits are finite and unequal. The signed jump is \(3-(-2)=5\), and the magnitude is \(5\). Although \(h(0)=7\), the value at \(0\) does not enter either limit calculation.
If \(h(0)\) is reassigned to \(-2\), the left-hand limit remains \(-2\) and the right-hand limit remains \(3\). Since those limits are still unequal, \(h\) is still discontinuous at \(0\). The reassignment changes one function value, not the values on either side.
Check: For every \(x<0\), \(|h(x)-(-2)|=0\), and for every \(x>0\), \(|h(x)-3|=0\). These differences are already zero on their respective sides, confirming the two limits.
A Common Mistake: Using the Point Value as a Limit
A frequent error is to inspect \(f(c)\) and treat it as though it determines what happens as \(x\) approaches \(c\). For example, in the function \(h\) above, an incorrect calculation might say: $$ \lim_{x\to0}h(x)=h(0)=7. $$ This is wrong. A two-sided limit can exist only if the one-sided limits agree, and here they are \(-2\) and \(3\). The point value \(7\) is not either limit and cannot make them agree.
The correction is to calculate each side separately, using the formula valid on that side: $$ \lim_{x\to0^-}h(x)=-2, \qquad \lim_{x\to0^+}h(x)=3. $$ Because \(-2\ne3\), the two-sided limit does not exist, and the discontinuity is a finite jump. The correct order is to find the one-sided limits first, compare them, and only then consider the value at the breakpoint.
How to Classify a Finite Jump
For a piecewise function with an interior breakpoint, a reliable test is:
Record the input \(c\) where the defining formula changes.
Calculate \(\lim_{x\to c^-}f(x)\), using only the branch that applies when \(x<c\).
Calculate \(\lim_{x\to c^+}f(x)\), using only the branch that applies when \(x>c\).
If both limits are finite and unequal, there is a finite jump. No reassignment of \(f(c)\) can make the function continuous at \(c\).
Graphically, the two sides approach different heights as the input approaches the breakpoint. A filled point at the breakpoint records \(f(c)\); it does not change the height approached from either side. This distinction is useful when reading graphs, analyzing piecewise formulas, or deciding whether a proposed change to a function can repair a discontinuity.
Check Your Understanding
For each question, keep the left-hand limit, the right-hand limit, and the function value distinct.
- What two conditions on the one-sided limits define a finite jump discontinuity?
- If the left-hand limit at \(c\) is \(4\) and the right-hand limit is \(-1\), what is the signed jump and its magnitude?
- For \(p(x)=2\) when \(x<3\), \(p(3)=10\), and \(p(x)=5\) when \(x>3\), calculate both one-sided limits and classify the discontinuity.
- Why does changing only \(f(c)\) leave the one-sided limits unchanged?
- If both one-sided limits are \(6\), does that fact alone guarantee continuity at \(c\)? What additional value condition is required?
