Why the Formula Depends on the Side
A piecewise function can follow one rule to the left of a breakpoint and a different rule to the right. To find a one-sided limit there, the first task is not to substitute the breakpoint into whichever formula is easiest to see. It is to identify the formula that actually describes the function on the side from which the input approaches.
For example, if a definition uses one formula for \(x<2\) and another for \(x\ge 2\), then values just below \(2\) are governed by the first formula. Values just above \(2\) are governed by the second. The fact that the second formula also assigns \(f(2)\) does not make it the formula for approaching from the left.
This distinction matters even when the function has a separately specified value at the breakpoint. A one-sided limit describes nearby values, not the value at the point itself.
Choosing the Branch
The superscript \(-\) indicates approach from the left, and \(+\) indicates approach from the right. In each case, \(x\) approaches \(c\), but the allowed values of \(x\) lie on a specified side. This is the one-sided limit notation used in the earlier material on one-sided limits.
A branch’s endpoint condition, such as \(x\le c\) or \(x<c\), determines whether that branch defines the function at \(c\). It does not change which branch describes nearby inputs on a given side. For a left-hand limit, select the rule valid immediately to the left; for a right-hand limit, select the rule valid immediately to the right.
Identify the input \(c\) toward which \(x\) approaches.
For \(x\to c^-\), consider inputs just less than \(c\); for \(x\to c^+\), consider inputs just greater than \(c\).
Read the piecewise definition on that side, without using the other branch just because it includes \(c\).
If the branch has a finite limit found by direct substitution, substitute \(c\) into that branch formula. If substitution does not determine the limit, use appropriate algebraic limit methods instead.
Why This Procedure Works
The key fact is that a branch formula agrees with the function throughout a small interval on its side of the breakpoint. Therefore, as the input approaches the breakpoint from that side, the function has exactly the same nearby values as that branch.
Here is a complete justification for the left-hand statement. By the definition of \(\lim_{x\to c^-}g_-(x)=L\), for every \(\varepsilon>0\) there is some \(\eta>0\) such that if \(c-\eta<x<c\), then \(|g_-(x)-L|<\varepsilon\). Choose a positive number \(\delta_0\) no greater than both \(\eta\) and \(\delta\). Whenever \(c-\delta_0<x<c\), the branch agreement gives \(f(x)=g_-(x)\), and the limit condition gives \(|g_-(x)-L|<\varepsilon\). Consequently, \(|f(x)-L|=|g_-(x)-L|<\varepsilon\). This is precisely the definition of \(\lim_{x\to c^-}f(x)=L\). The right-hand statement follows by the same reasoning with \(c<x<c+\delta_0\).
This argument also explains why the assigned value \(f(c)\) does not determine either one-sided limit: the definition and proof use inputs strictly to one side of \(c\), not the input \(x=c\).
Worked Examples
Worked Example: Different Rules on Either Side
Let \(f\) be defined by
Find \(\lim_{x\to 2^-}f(x)\) and \(\lim_{x\to 2^+}f(x)\).
For \(x\to2^-\), inputs are less than \(2\), so use \(x^2+1\). Substituting \(2\) into this polynomial gives
For \(x\to2^+\), inputs are greater than \(2\), so use \(4x-1\). Thus
The two one-sided limits differ: the left-hand limit is \(5\), and the right-hand limit is \(7\). Check the branch selection: the left value comes from the formula explicitly valid for \(x<2\), and the right value comes from the formula valid for \(x\ge2\) when \(x>2\).
Worked Example: Simplify the Branch Before Substitution
Define
Find both one-sided limits at \(x=2\). On the left, the relevant expression is a quotient that gives \(0/0\) if substituted directly. Factor its numerator first:
The cancellation is valid for the inputs in this branch because \(x<2\), so \(x-2\ne0\). Therefore,
On the right, use the branch \(x+2\), which applies for \(x\ge2\), including inputs just above \(2\):
Both limits are \(4\). As a check, on the left the original quotient equals \(x+2\) for every permitted \(x<2\), and on the right the function is already \(x+2\). Both branch behaviors therefore approach \(4\).
Worked Example: A Breakpoint Inside a Multi-Branch Definition
Suppose
Find the left- and right-hand limits at \(x=3\). Immediately to the left of \(3\), inputs satisfy \(0\le x<3\), so the middle formula applies. Hence
Immediately to the right of \(3\), inputs satisfy \(x\ge3\), so the last formula applies. Therefore,
The first formula, \(2-x\), is not relevant to either limit at \(3\): it applies only for \(x<0\). The left and right answers are \(10\) and \(4\), respectively. This check confirms that the selected intervals meet the breakpoint \(3\) on the correct sides.
The Function’s Value Is a Separate Question
A piecewise definition can specify \(f(c)\) through a branch that includes equality, but that value is separate from the one-sided limits. For instance, in the first example the definition assigns \(f(2)=4(2)-1=7\). The left-hand limit is nevertheless \(5\), because values below \(2\) follow \(x^2+1\). The right-hand limit is \(7\), since values just above \(2\) follow \(4x-1\).
In general, finding a limit means examining values near the input, while evaluating the function means using its definition at that input. The function’s value might agree with one of the one-sided limits, with both, or with neither. Do not infer a limit solely from the branch that assigns the value at the breakpoint.
If both one-sided limits exist as finite numbers, they are equal exactly when the two side-specific values agree. A difference between them means that the function approaches different values from the two sides. This conclusion comes from comparing the separately calculated limits, not from inspecting \(f(c)\).
Common Pitfall: Using the Branch That Includes the Breakpoint
A frequent error is to use the formula containing equality for both one-sided limits. For the first example, that mistake would give
This uses the right branch, even though \(x\to2^-\) means that \(x<2\). The correction is to select \(x^2+1\) for the left-hand limit:
The equality condition decides which formula defines the value at the breakpoint. The direction of approach decides which formula describes nearby values. Keeping those two questions separate prevents the error.
When Direct Substitution Is Not Enough
Once the correct branch is selected, direct substitution gives the limit when that branch’s expression has a limit determined by its ordinary algebraic behavior at \(c\), as with the polynomials in the examples. But selecting the branch does not guarantee that substitution will immediately produce an answer. In the second example, substituting into the left branch first produced \(0/0\); factoring was needed before evaluating the limit.
The order of reasoning remains the same: determine the side, choose its branch, and then evaluate that branch’s limit using an applicable limit method. Algebraic laws for one-sided limits and earlier methods such as factoring can help with the last step. A calculation on the wrong branch cannot be repaired by correct algebra.
Check Your Understanding
For each question, first identify the formula valid on the specified side.
- If \(p(x)=x+1\) for \(x<4\) and \(p(x)=2x\) for \(x\ge4\), which formula should be used to find \(\lim_{x\to4^-}p(x)\), and what is the limit?
- For the same function, which formula should be used to find \(\lim_{x\to4^+}p(x)\), and what is the limit?
- Why does the value assigned to \(p(4)\) not by itself determine \(\lim_{x\to4^-}p(x)\)?
- If a branch simplifies to \(x+5\) for every \(x<1\), what is its limit as \(x\to1^-\)?
- At a breakpoint, what does it mean if the two one-sided limits exist but are unequal?
