Why the Branch and Its Sign Matter
A piecewise function can use one formula to the left of a point and a different formula to the right. Its two one-sided limits must therefore be calculated separately, using the formula that actually applies on each side.
For rational expressions, factoring is often the first useful step. It can reveal a common factor that simplifies to a finite limit, or show that a denominator approaches zero with a particular sign. That sign determines whether the quotient grows toward positive or negative infinity.
The value assigned to the function at the boundary is not what determines either one-sided limit. The limits describe nearby function values, and each side may follow a different rule.
One-Sided Limits at a Piecewise Boundary
A one-sided limit may be a finite number, \(+\infty\), or \(-\infty\). The notation \(\lim_{x\to a^+}f(x)=+\infty\) means that the function values grow without bound in the positive direction as \(x\) approaches \(a\) from the right. The other infinite-limit notation has the corresponding meaning for values decreasing without bound.
Factoring helps expose both the size and sign of a rational expression near the boundary. For example, if a factor \(x-a\) remains in the denominator and all other factors approach nonzero values, the denominator approaches zero. Whether it is positive or negative on the chosen side then determines the sign of the quotient, provided the numerator also has a definite nonzero sign.
A Sign-Based Principle
The reason is that \(u(x)\) eventually has the sign of \(L\) and stays bounded away from zero. For instance, if \(L>0\), then sufficiently near \(a\), \(u(x)>L/2>0\). If at the same time \(v(x)>0\) and \(v(x)\to0\), then for any \(M>0\), sufficiently near \(a\) we have \(0<v(x)<L/(2M)\). Therefore \(u(x)/v(x)>M\). Since this holds for every \(M>0\), the quotient tends to \(+\infty\). The other sign cases follow by reversing the signs of the numerator or denominator.
This principle explains why “the denominator tends to zero” is not enough to determine an infinite limit: its sign on the chosen side matters. If the numerator also tends to zero, first factor and simplify where the original expression is defined. A zero numerator and zero denominator alone do not determine the limit.
Worked Examples
Worked Example: A Finite Limit on One Side and an Infinite Limit on the Other
Let
Find \(\lim_{x\to1^-}f(x)\) and \(\lim_{x\to1^+}f(x)\).
For the left-hand limit, use the branch for \(x<1\). Factor the numerator:
The cancellation is valid for inputs in this branch because \(x\ne1\). Thus
For the right-hand limit, use the branch for \(x>1\). As \(x\to1^+\), the numerator \(x+2\to3\), which is positive near \(1\). The denominator \(x-1\to0\) through positive values because \(x>1\). Hence the quotient grows without bound in the positive direction:
Check the conclusions: the simplified left branch \(x+1\) approaches \(2\); on the right, a numerator near \(3>0\) divided by a small positive denominator is positive and arbitrarily large. The assigned value \(f(1)=7\) changes neither limit.
Worked Example: A Negative Infinite Limit on the Left and a Finite Limit on the Right
Let
Find the left-hand and right-hand limits at \(x=2\).
On the left, the numerator \(x+1\) approaches \(3\), so it is positive near \(2\). The factor \(x-2\) is negative for \(x<2\) and approaches zero. The other denominator factor \(x+3\) approaches \(5\), so it is positive near \(2\). Thus the denominator is a negative quantity approaching zero: it approaches \(0^-\). Consequently, the quotient tends to negative infinity:
On the right, factor the numerator and cancel the common factor for \(x>2\):
Therefore
Check: on the left, the numerator is positive and the denominator is negative and close to zero, giving a large negative quotient. On the right, the simplified branch approaches \(4\). Both results use the correct branch for their side.
Worked Example: Opposite Infinite Limits at the Same Boundary
Let
Find the one-sided limits at \(x=0\).
For \(x\to0^-\), the numerator \(x^2+1\) approaches \(1\) and is positive. The denominator contains \(x<0\) and \(x-1<0\) near zero. Their product is positive. Its magnitude approaches zero, since \(x\to0\) and \(x-1\to-1\). Thus the quotient is positive and grows without bound:
For \(x\to0^+\), use the other branch. The numerator \(x-1\) approaches \(-1\) and is negative near zero. The denominator has \(x>0\) and \(x+2>0\), so it is positive and approaches zero. A negative numerator divided by a small positive denominator tends to negative infinity:
Check: the left branch has a positive numerator and a positive denominator close to zero; the right branch has a negative numerator and a positive denominator close to zero. The opposite signs of the quotients agree with the two results. The value \(h(0)=5\) does not determine either limit.
A Reliable Procedure
At a piecewise boundary \(x=a\), handle each side independently. The following sequence keeps branch selection, factoring, and sign analysis distinct:
For \(x\to a^-\), use the formula specified for inputs less than \(a\). For \(x\to a^+\), use the formula specified for inputs greater than \(a\).
Look for common factors in the numerator and denominator. Cancel a common factor only for nearby inputs where the original expression is defined; use the resulting expression to evaluate the limit.
If it approaches a nonzero value, the quotient has a finite limit given by the limiting numerator divided by the limiting denominator.
Track each factor using whether \(x\) is just below or just above \(a\). Combine that sign with the numerator’s sign to decide between \(+\infty\) and \(-\infty\).
A left-hand limit and a right-hand limit can be finite, infinite, or different from one another. Do not infer one side from the other.
Common Sign and Branch Errors
A frequent error is to use the branch that includes the boundary value instead of the branch that applies to nearby inputs on the side being considered. For the function \(f\) in the first example, \(f(1)=7\), but the left-hand limit is \(2\) and the right-hand limit is \(+\infty\). The incorrect calculation “\(f(1)=7\), so both limits are \(7\)” confuses the value at a point with the behavior of nearby values.
Another common error is to notice a denominator approaching zero and immediately write \(+\infty\). In the second example, for \(x\to2^-\), the denominator \((x-2)(x+3)\) is negative and approaches zero. Since the numerator is positive near \(2\), the quotient tends to \(-\infty\), not \(+\infty\).
Why the Two Sides Must Be Kept Separate
Piecewise definitions can change their formulas at the boundary, and even one formula can behave differently on opposite sides because a factor such as \(x-a\) changes sign there. As a result, the two one-sided limits may be unequal, or one may be finite while the other is infinite. These conclusions come from the formulas near the boundary, not from the function’s assigned value at the boundary.
Working one side at a time also makes the method easier to check. The branch identifies the expression; factoring reveals its structure; sign analysis determines the direction of any unbounded behavior. Each step answers a separate question, and together they establish the correct one-sided limit.
Check Your Understanding
For each question, select the relevant branch before calculating. State the sign reasoning when a denominator approaches zero.
- Let \(p(x)=\frac{x^2-9}{x-3}\) for \(x<3\), and \(p(x)=\frac{x+1}{x-3}\) for \(x>3\). Find \(\lim_{x\to3^-}p(x)\) and \(\lim_{x\to3^+}p(x)\).
- For \(q(x)=\frac{x+2}{(x+1)(x-1)}\) when \(x<1\), determine \(\lim_{x\to1^-}q(x)\). What are the signs of the numerator and denominator near \(1\) on that side?
- For \(r(x)=\frac{x^2-4}{x-2}\) when \(x>2\), factor and find \(\lim_{x\to2^+}r(x)\). Why is cancellation valid for this limit?
- Can the value assigned to a piecewise function at \(x=a\) determine either one-sided limit? Explain briefly.
- If a rational expression has a numerator approaching a negative nonzero number and a denominator approaching zero through positive values, which infinite limit does it have?
