Why Several Breakpoints Need an Organized Check
A piecewise function can change its defining formula at more than one input. Each change creates a point that must be checked separately. Checking one breakpoint carefully does not establish continuity at the others, even if the same formulas appear again.
The reliable approach is to list the breakpoints in order and, at each one, compare three things: the limit from the left, the limit from the right, and the value of the function at the breakpoint. This makes it clear both where continuity holds and, when it fails, what caused the failure.
What Counts as a Breakpoint?
A breakpoint is a point where the rule defining a piecewise function changes. For example, if a function uses one formula for \(x<0\), another for \(0\le x<1\), and a third for \(x\ge1\), the breakpoints are \(0\) and \(1\). The inequalities determine which formula applies on each side and which formula gives the function value at the breakpoint.
At a breakpoint \(c\), the formula used for \(x<c\) determines the left-hand limit, and the formula used for \(x>c\) determines the right-hand limit. The formula whose interval includes \(c\), or a separately stated value \(f(c)\), determines the function value. A formula’s presence on one side does not by itself determine \(f(c)\).
When each formula is continuous near its relevant breakpoint, its one-sided limit can be found by substituting the breakpoint into that formula. If a formula is not continuous there, its one-sided limit must instead be evaluated using the limit techniques appropriate to that formula. The check must include every breakpoint; formulas that repeat still have to be evaluated at the new point.
Why the Three-Part Test Works
Continuity at \(c\) means that a single two-sided limit exists and equals \(f(c)\). The two-sided limit exists exactly when the left-hand and right-hand limits exist and have the same value. Therefore, continuity requires those two limits to agree with each other and with the function’s assigned value.
To see why the one-sided comparison captures the two-sided limit, suppose both one-sided limits equal \(L\). Given any \(\varepsilon>0\), the definition of the left-hand limit gives a positive distance \(\delta_-\) such that \(0<c-x<\delta_-\) implies \(|f(x)-L|<\varepsilon\). The right-hand limit gives a positive distance \(\delta_+\) such that \(0<x-c<\delta_+\) implies the same inequality. Choose \(\delta=\min(\delta_-,\delta_+)\). Then \(0<|x-c|<\delta\) places \(x\) on one side or the other, and in either case \(|f(x)-L|<\varepsilon\). Thus \(\lim_{x\to c}f(x)=L\).
Conversely, if the two-sided limit is \(L\), its defining inequality holds for all \(x\) sufficiently close to \(c\), with \(x\ne c\). Restricting those \(x\)-values to the left or to the right proves that both one-sided limits equal \(L\). This justifies checking the one-sided limits separately at each breakpoint.
A Systematic Procedure
For a piecewise function with several breakpoints, use the following process. It keeps the calculation local: each breakpoint gets its own three entries, even if a formula has appeared earlier.
Read them from the interval conditions. Check that each listed point is in the domain and note which formula applies on each side.
Use the formula on the interval immediately to the left of the breakpoint and evaluate its limit from that side.
Use the formula on the interval immediately to the right and evaluate its limit from that side.
Use the piece of the definition that includes the breakpoint, or the separately assigned value if one is given.
The function is continuous at this point precisely when the left-hand limit, right-hand limit, and function value are equal. Repeat the test at every breakpoint.
Worked Example: Checking Three Breakpoints with Repeated Formulas
Worked Example: Checking Three Breakpoints with Repeated Formulas
Consider the function
The breakpoints are \(0\), \(1\), and \(2\). The formula \(x+1\) occurs on two separate intervals, so evaluate it at the relevant breakpoint each time it supplies a one-sided limit.
At \(x=0\), the left-hand formula is \(x+1\), so \(L_-=\lim_{x\to0^-}(x+1)=0+1=1\). The right-hand formula is \(x^2+1\), so \(L_+=\lim_{x\to0^+}(x^2+1)=0^2+1=1\). Since \(0\) belongs to the second piece, \(f(0)=0^2+1=1\). Thus \(L_-=L_+=f(0)=1\), and \(f\) is continuous at \(0\).
At \(x=1\), the left-hand formula is \(x^2+1\), so \(L_-=\lim_{x\to1^-}(x^2+1)=1^2+1=2\). The right-hand formula is \(x+1\), so \(L_+=\lim_{x\to1^+}(x+1)=1+1=2\). Since \(1\) belongs to the third piece, \(f(1)=1+1=2\). Therefore \(L_-=L_+=f(1)=2\), and \(f\) is continuous at \(1\).
At \(x=2\), the left-hand formula is \(x+1\), so \(L_-=\lim_{x\to2^-}(x+1)=2+1=3\). The right-hand formula is \(3x-3\), so \(L_+=\lim_{x\to2^+}(3x-3)=3(2)-3=6-3=3\). Since \(2\) belongs to the fourth piece, \(f(2)=3(2)-3=3\). Thus \(L_-=L_+=f(2)=3\), and \(f\) is continuous at \(2\).
The function is continuous at all three breakpoints. Each formula is a polynomial and is continuous wherever it applies, so there are no additional failures within the pieces. The breakpoint checks are therefore complete.
Worked Example: One Failed Breakpoint Among Several
Worked Example: One Failed Breakpoint Among Several
Consider
The breakpoints are \(-1\) and \(1\). At \(x=-1\), the left-hand limit is \(L_-=\lim_{x\to-1^-}x^2=(-1)^2=1\). The right-hand limit is \(L_+=\lim_{x\to-1^+}(2x+4)=2(-1)+4=-2+4=2\). The point \(-1\) belongs to the middle piece, so \(g(-1)=2(-1)+4=2\). Since \(L_-=1\) but \(L_+=g(-1)=2\), the two one-sided limits do not agree. Thus \(g\) is not continuous at \(-1\).
At \(x=1\), the left-hand limit is \(L_-=\lim_{x\to1^-}(2x+4)=2(1)+4=6\). The right-hand limit is \(L_+=\lim_{x\to1^+}(x+5)=1+5=6\). Since \(1\) belongs to the third piece, \(g(1)=1+5=6\). Hence \(L_-=L_+=g(1)=6\), and \(g\) is continuous at \(1\).
Each piece is a polynomial, so it is continuous within its interval. The complete conclusion is that \(g\) fails to be continuous at \(-1\), but is continuous at \(1\). A successful check at \(1\) does not repair the failure at \(-1\).
Worked Example: Matching Limits but a Different Point Value
Worked Example: Matching Limits but a Different Point Value
Define
At \(x=0\), the left-hand limit is \(L_-=\lim_{x\to0^-}(x+1)=0+1=1\). The right-hand limit is \(L_+=\lim_{x\to0^+}(x^2+1)=0^2+1=1\). But the definition assigns \(h(0)=7\). Thus the two one-sided limits agree with each other, but \(L_-=L_+=1\ne7=h(0)\). The function is not continuous at \(0\).
At \(x=1\), the left-hand limit is \(L_-=\lim_{x\to1^-}(x^2+1)=1^2+1=2\). The right-hand limit is \(L_+=\lim_{x\to1^+}2x=2(1)=2\). The value is \(h(1)=2(1)=2\), because the final piece includes \(1\). Therefore \(L_-=L_+=h(1)=2\), and \(h\) is continuous at \(1\).
This check shows why agreement of the two one-sided limits alone is not enough: their common value must also equal the function value.
Common Pitfall: Checking Only the Formula on the Right
At a breakpoint, substituting into the formula that includes the point gives the function value, but it does not establish continuity. For example, in the preceding function \(h\), the point value is \(h(0)=7\). A mistaken check might stop there and report that the function is defined at \(0\), or might compare \(h(0)\) only with the right-hand expression. But the actual right-hand limit is \(1\), and the left-hand limit is also \(1\), so neither equals \(h(0)\).
The correction is to keep three separate entries for every breakpoint: left-hand limit, right-hand limit, and function value. A function can fail the test because the one-sided limits differ, as in \(g\) at \(-1\), or because the limits agree but the assigned value differs, as in \(h\) at \(0\).
Why the Breakpoint Table Is Useful
For a function with many pieces, recording the calculations in a table helps prevent omissions. The table is not a substitute for evaluating the limits; it is a way to make sure that each breakpoint receives the complete test.
| Breakpoint | Left-hand limit | Right-hand limit | Function value | Conclusion |
|---|---|---|---|---|
| \(c\) | From the formula to the left | From the formula to the right | From the definition at \(c\) | Continuous exactly when all three agree |
If a function has breakpoints \(c_1<c_2<\cdots<c_n\), apply the test at \(c_1\), then \(c_2\), and continue through \(c_n\). Do not infer a result at a later point from an earlier one, even when the same formula is used again. If the domain has an endpoint rather than an interval on both sides, the endpoint test uses the one-sided limit from within the domain; at an interior breakpoint, both sides must be checked.
Once every breakpoint has been checked, also account for continuity within the pieces. If each piece is continuous on its interval and the three-part test succeeds at every interior breakpoint, then the function is continuous throughout the combined interval. If a formula has its own discontinuity inside a piece, that point needs its own limit check as well.
Check Your Understanding
For each question, organize the relevant limits and function values before drawing a conclusion.
- For \(p(x)=x+2\) when \(x<0\), \(p(x)=x^2+2\) when \(0\le x<1\), and \(p(x)=2x+1\) when \(x\ge1\), identify every breakpoint and test continuity at each.
- At an interior breakpoint, what three quantities must agree for continuity?
- Why must a repeated formula still be evaluated at each breakpoint where it supplies a one-sided limit?
- A function has equal left-hand and right-hand limits at \(c\), but \(f(c)\) differs from their common value. Is it continuous at \(c\)? Explain using the continuity test.
- If each formula is continuous on its interval and all breakpoint tests succeed, what can be concluded about continuity across the combined interval?
