One sided limits and when a limit exists
A function can head toward one height from the left and toward another from the right. When does it have a limit at all, and how do you find each side from a formula rather than a picture?
Where you are 11 in Limits
Why this exists
In What a limit is: graphs and tables a limit was read off a picture, one side at a time, using the one-sided limits \lim_{x \to a^-} f(x) and \lim_{x \to a^+} f(x); that lesson stated without proof that the limit exists exactly when they agree.
Many functions are built from different rules on different sides of a point: a piecewise function, a quotient with an absolute value, a step function, a square root that stops at the edge of its domain. For these, a limit is found by splitting the question in two, settling each side from the formula that governs it, and comparing. This lesson proves the splitting is legitimate and carries it out with formulas.
The lesson takes two days. Part 1 is the criterion itself, why it is true, and one-sided limits of piecewise functions found from their rules; Examples 1 to 4 go with it. Part 2 is absolute-value quotients, the greatest integer function, and limits at an endpoint of a domain; Examples 5 to 7 go with it.
Definition
The left-hand limit of f at a is L, written \lim_{x \to a^-} f(x) = L, when f(x) is as close as you like to L for every input x that is less than a and close enough to a. The right-hand limit \lim_{x \to a^+} f(x) = M is the same with inputs greater than a. Each one asks about one side only, and neither ever looks at the input a itself.
The two-sided limit \lim_{x \to a} f(x) = L asks that f(x) be as close as you like to L for every x \ne a close enough to a, on either side. It needs f to be defined at every input near a on both sides, except perhaps at a.
When the outputs do not approach one number, the limit does not exist. Writing \lim_{x \to a^+} f(x) = \infty records that the outputs grow without bound; the limit still does not exist as a number.
Intuition
The two-sided limit is two requests at once: the left side must settle on a height, the right side must settle on a height, and the heights must be the same. If both sides head to 1, as in Figure 1, every input near a gives an output near 1. If the left heads to -7 and the right to 7, as in Figure 3, there is no single number for the outputs to crowd around. The point above a is never asked about, so moving, removing or adding the dot at x = a changes nothing.
With formulas, one idea does most of the work: a one-sided limit sees only the rule on its own side. If f(x) = x^2 - 3 for every x just left of 2, the left-hand limit of f at 2 is that of x^2 - 3, whatever f does at 2 or to the right, and for a polynomial that is its value, 2^2 - 3 = 1. The rule may be evaluated at 2 even though its condition x < 2 leaves 2 out: the result is not f(2), it is the height the left side heads toward.
Results
Theorem (the two-sided limit criterion). Let f be defined at every input near a on both sides, except perhaps at a. Then
Consequences, each used below:
- If either one-sided limit does not exist (including one that runs off to \infty or -\infty), the two-sided limit does not exist.
- If both one-sided limits exist but are different numbers, the two-sided limit does not exist.
- The value f(a), and whether f(a) is defined at all, has no effect on any of the three limits.
- Locality. If f(x) = g(x) for every x in some interval a - d < x < a, then f and g have the same left-hand limit at a (or both have none). The same holds on the right.
- For a polynomial g, and for square roots and quotients of polynomials at inputs inside their domains, \lim_{x \to a} g(x) = g(a). This is proved with the limit laws and is taken here without proof.
At an endpoint of a domain. If f is defined only on one side of a, as \displaystyle \sqrt{x} is at 0, the two-sided limit is not asked; only the one-sided limit from inside the domain is meaningful: \displaystyle \lim_{x \to 0^+} \sqrt{x} = 0.
Why it is true
The argument uses only the meaning of “as close as you like”. Call the demanded closeness a tolerance: a positive number that |f(x) - L| must stay below.
Two-sided gives both one-sided. Suppose \lim_{x \to a} f(x) = L and a tolerance is named. Then there is a distance d so that every x \ne a with |x - a| < d gives |f(x) - L| below the tolerance. The inputs in that window that are less than a are some of those inputs, so they meet the tolerance too. Since this works for every tolerance, \lim_{x \to a^-} f(x) = L. The same reasoning with the inputs greater than a gives \lim_{x \to a^+} f(x) = L.
Both one-sided give two-sided. Suppose both one-sided limits equal L and a tolerance is named. The left-hand limit supplies a distance d_1 so that every x with a - d_1 < x < a meets the tolerance; the right-hand limit supplies d_2 for a < x < a + d_2. Let d be the smaller of d_1 and d_2. An input x \ne a with |x - a| < d lies either just left of a, inside the first window, or just right of a, inside the second, so it meets the tolerance either way. Hence \lim_{x \to a} f(x) = L.
When the sides fail or disagree. If the two-sided limit existed and equaled some L, the first part would force both one-sided limits to exist and to equal that same L. So if one side has no limit, or the two sides head to different numbers, no two-sided limit can exist. (A one-sided limit, like a two-sided one, is a single number when it exists: outputs cannot stay close to two numbers a fixed distance apart.)
No step looked at x = a, so f(a) plays no part. Locality holds because a one-sided limit uses only inputs within some small distance of a on its side, where f and g agree. With the tolerance written \varepsilon and the distance \delta, this is a proof in the sense of The epsilon delta definition of a limit.
Worked examples
Ingvild’s luggage fare
A ferry between two islands of the Faroes charges for checked luggage by weight w in kilograms:
in dollars (Figure 2). Ingvild’s bag weighs close to 20 kilograms. Find \lim_{w \to 20^-} F(w), \lim_{w \to 20^+} F(w), \lim_{w \to 20} F(w) and F(20). Then find \lim_{w \to 12} F(w).
For weights between 19 and 20 the first rule gives 12 every time, and outputs that all equal 12 are as close as you like to 12, so \lim_{w \to 20^-} F(w) = 12. For weights between 20 and 21 every output is 18, so \lim_{w \to 20^+} F(w) = 18.
The one-sided limits are different numbers, so by the criterion \lim_{w \to 20} F(w) does not exist. The condition 0 \le w \le 20 includes 20, so F(20) = 12. The fare at 20 kilograms is well defined; what fails is a single number that fares near 20 kilograms crowd around.
At w = 12, every weight between 11 and 13 falls under the first rule, so both one-sided limits are 12 and \lim_{w \to 12} F(w) = 12.
Both rules arriving at one point
Let
Find both one-sided limits at 2, decide whether \lim_{x \to 2} h(x) exists, and compare it with h(2). Also find \lim_{x \to -1} h(x).
SolutionFor x just less than 2, h(x) = x^2 - 3. By locality the left-hand limit of h is the left-hand limit of the polynomial x^2 - 3, and a polynomial’s outputs near 2 are near its value there:
For x just greater than 2, h(x) = 5 - 2x, so
Both one-sided limits exist and equal 1, so \lim_{x \to 2} h(x) = 1. The condition x \ge 2 puts 2 in the second rule, so h(2) = 5 - 2 \cdot 2 = 1. The two pieces of Figure 1 meet at (2, 1).
Near x = -1 only the first rule is in force on both sides, so \lim_{x \to -1} h(x) = (-1)^2 - 3 = -2.
Sjurdur’s function with one side that runs away
Let
Sjurdur finds \lim_{x \to -1^-} q(x) = -1 and stops there. Does \lim_{x \to -1} q(x) exist?
SolutionHis left-hand limit is right: for x < -1 the rule is the polynomial x^2 + 2x, so \lim_{x \to -1^-} q(x) = (-1)^2 + 2(-1) = -1. For instance q(-1.01) = -0.9999.
The right side uses \displaystyle \frac{1}{x + 1}, where -1 cannot be substituted. For x just greater than -1, x + 1 is small and positive, so its reciprocal is large: \displaystyle q(-0.99) = \frac{1}{0.01} = 100 and \displaystyle q(-0.999) = \frac{1}{0.001} = 1000. The outputs pass every number, so
and the right-hand limit does not exist as a number. The criterion needs both sides, so \lim_{x \to -1} q(x) does not exist, even though the left side settles on -1. That q(-1) is undefined would not have mattered either way. A side that runs off to infinity is the subject of Infinite limits and vertical asymptotes.
Every boundary of a three-rule function
Let
Find every input a at which \lim_{x \to a} r(x) does not exist.
SolutionAny input other than 0 and 3 has a small window around it that uses a single polynomial rule, so by locality the limit there exists and is that polynomial’s value. Only the two boundaries need checking.
At a = 0: from the left, \lim_{x \to 0^-} (2 - x) = 2; from the right, \lim_{x \to 0^+} (x^2 + 2) = 2. The sides agree, so \lim_{x \to 0} r(x) = 2, and r(0) = 0^2 + 2 = 2 as well.
At a = 3: from the left, \lim_{x \to 3^-} (x^2 + 2) = 9 + 2 = 11; from the right, \lim_{x \to 3^+} (x + 5) = 8. The sides disagree, so \lim_{x \to 3} r(x) does not exist, though r(3) = 3 + 5 = 8 is defined and matches the right side; the criterion never consults r(3).
So the limit fails at exactly one input, a = 3.
An absolute value in the denominator
Let s(x) = \dfrac{x^2 - 3x - 10}{|x - 5|}. Find \lim_{x \to 5^-} s(x), \lim_{x \to 5^+} s(x) and \lim_{x \to 5} s(x). Then find \lim_{x \to 7} s(x) and \lim_{x \to 2} s(x).
SolutionAn absolute value is a piecewise rule, |u| = u when u \ge 0 and |u| = -u when u < 0, so s is piecewise with boundary x = 5, where x - 5 changes sign. The top factors as (x - 5)(x + 2).
For x < 5, x - 5 is negative, so |x - 5| = -(x - 5) and
the cancellation being allowed because x - 5 \ne 0. For x > 5, |x - 5| = x - 5 and s(x) = x + 2. Each side is now a polynomial rule, so
The sides disagree, so \lim_{x \to 5} s(x) does not exist, and s(5) is undefined since the denominator is 0 (Figure 3).
Near 7 every input exceeds 5, so s(x) = x + 2 on a whole window around 7 and \lim_{x \to 7} s(x) = 9. Near 2 every input is below 5, so \lim_{x \to 2} s(x) = -(2 + 2) = -4.
Joannes and the part after the decimal point
The greatest integer function \lfloor x \rfloor gives the largest integer less than or equal to x. Joannes studies g(x) = x - \lfloor x \rfloor, the amount by which x exceeds the integer below it. Find \lim_{x \to 2^-} g(x), \lim_{x \to 2^+} g(x) and g(2), decide whether \lim_{x \to 2} g(x) exists, and find \lim_{x \to 2.5} g(x).
SolutionThe floor function is a step function, constant between consecutive integers: \lfloor x \rfloor = 1 for 1 \le x < 2 and \lfloor x \rfloor = 2 for 2 \le x < 3. So near 2,
From the left, \lim_{x \to 2^-} (x - 1) = 1. From the right, \lim_{x \to 2^+} (x - 2) = 0. The sides disagree, so \lim_{x \to 2} g(x) does not exist, while g(2) = 2 - 2 = 0. For example g(1.99) = 0.99 and g(2.01) = 0.01: two inputs 0.02 apart give outputs almost 1 apart.
At every integer the left side heads to 1 and the right side to 0 (Figure 4). At 2.5 the window 2 < x < 3 uses the single rule x - 2, so \lim_{x \to 2.5} g(x) = 0.5 = g(2.5). The floor function itself, at an integer n, has left-hand limit n - 1 and right-hand limit n.
Where the domain stops
Find \displaystyle \lim_{x \to 1^+} \sqrt{x - 1}, and explain why no two-sided limit is asked at 1. Then let
and decide whether \lim_{x \to 3} b(x) exists.
SolutionThe root \displaystyle \sqrt{x - 1} is defined only for x \ge 1, so there are no inputs left of 1, and the two-sided limit, which needs both sides, is not asked. For the right-hand limit, name a tolerance t > 0. Then \displaystyle \sqrt{x - 1} < t exactly when 0 \le x - 1 < t^2, so every x between 1 and 1 + t^2 gives an output within t of 0. Hence
For instance \displaystyle \sqrt{1.01 - 1} = 0.1 and \displaystyle \sqrt{1.0001 - 1} = 0.01. Some books define the limit using only inputs in the domain and would write \displaystyle \lim_{x \to 1} \sqrt{x - 1} = 0; the one-sided limit says the same without depending on that choice. The fare of Example 1 is another case: at w = 0 only \lim_{w \to 0^+} F(w) = 12 is meaningful.
The function b is different. Each rule alone stops at 3, but together they cover both sides. From the left \displaystyle \sqrt{3 - x} heads to 0 by the same tolerance argument, and from the right \displaystyle \sqrt{x - 3} heads to 0, so \lim_{x \to 3} b(x) = 0, and b(3) = 0. Whether a two-sided limit is asked depends on whether the function has inputs on both sides, not on whether each rule does.
Boundaries
- The criterion needs f defined on both sides near a. At an endpoint of the domain only the one-sided limit from the inside is meaningful (Example 7).
- A one-sided limit equal to \infty or -\infty is a way of not existing. If both sides go to \infty, writing \lim_{x \to a} f(x) = \infty describes the failure; the limit still does not exist as a number.
- Substituting a into a piece’s rule gives that side’s limit only when the rule’s limit at a is its value there, as for a polynomial. A rule like \displaystyle \frac{1}{x + 1} at -1 (Example 3) cannot be handled this way, and a \displaystyle \frac{0}{0} rule needs Limits by algebra: factoring and rationalizing.
- Locality needs a whole window of inputs on one side; values at a few inputs near a decide nothing.
- Equal one-sided limits say nothing about f(a), and f(a) says nothing about either one-sided limit.
Where this goes next
The criterion is the working tool of Continuity: a piecewise function is continuous at a boundary exactly when both one-sided limits exist, agree, and equal the value, and the jumps of Examples 1, 4, 5 and 6 become a named kind of break. One-sided limits that run off to infinity, as in Example 3, are the language of Infinite limits and vertical asymptotes. The limit laws prove that a polynomial’s limit is its value.
Exercises
Part 1: one-sided limits from the rules
- For the fare F of Example 1, find \lim_{w \to 30^-} F(w), \lim_{w \to 30^+} F(w), \lim_{w \to 30} F(w) and F(30).
- For the same fare, find \lim_{w \to 35} F(w) and \lim_{w \to 26} F(w).
- Let f(x) = x^2 + 1 for x < 3 and f(x) = 4x - 2 for x \ge 3. Find both one-sided limits at 3, the limit, and f(3).
- Let g(x) = 3 - x for x \le -2 and g(x) = x^2 + 2x for x > -2. Find both one-sided limits at -2, the limit, and g(-2).
- Let m(x) = x^3 for x < 1, m(1) = 5, and m(x) = 2 - x^2 for x > 1. Find both one-sided limits at 1, the limit, and m(1).
- Let p(x) = 2x - 5 for x < 4 and p(x) = \dfrac{1}{x - 4} for x > 4. Find \lim_{x \to 4^-} p(x), describe p(x) as x \to 4^+, and decide whether \lim_{x \to 4} p(x) exists.
- Let u(x) = x + 4 for x < -1, u(x) = x^2 + 2 for -1 \le x \le 2, and u(x) = 10 - x for x > 2. Find \lim_{x \to -1} u(x), \lim_{x \to 0} u(x), and decide whether \lim_{x \to 2} u(x) exists.
- Let \displaystyle v(x) = \sqrt{x + 5} for x < 4 and v(x) = x - 1 for x \ge 4. Find both one-sided limits at 4, the limit, and v(4).
Part 2: absolute values, floors and endpoints
- Find \lim_{x \to 3^-} \dfrac{|2x - 6|}{x - 3}, \lim_{x \to 3^+} \dfrac{|2x - 6|}{x - 3}, and decide whether the two-sided limit exists.
- Find both one-sided limits of \dfrac{x^2 - x}{|x - 1|} at 1, and decide whether the limit exists.
- Find \lim_{x \to 4} \dfrac{|x - 1|}{x - 1} and \lim_{x \to -3} \dfrac{|x - 1|}{x - 1}.
- Find \lim_{x \to 3^-} (x + |x - 3|), \lim_{x \to 3^+} (x + |x - 3|) and \lim_{x \to 3} (x + |x - 3|).
- Find \lim_{x \to -2^-} \lfloor x \rfloor, \lim_{x \to -2^+} \lfloor x \rfloor, and \lim_{x \to -1.5} \lfloor x \rfloor.
- Find both one-sided limits of \displaystyle \left\lfloor \frac{x}{2} \right\rfloor at 4, and find \displaystyle \lim_{x \to 5} \left\lfloor \frac{x}{2} \right\rfloor.
- Let k(x) = \lfloor x \rfloor + \lfloor -x \rfloor. Find \lim_{x \to 3^-} k(x), \lim_{x \to 3^+} k(x), \lim_{x \to 3} k(x) and k(3).
- For the fare F of Example 1, find \lim_{w \to 40^-} F(w), and say why no right-hand limit is asked at 40.
- Find \displaystyle \lim_{x \to 9^-} \sqrt{9 - x}, and say which one-sided limit is meaningful at 9 and why.
- Find \displaystyle \lim_{x \to 2^+} \left(x + \sqrt{x - 2}\right).
- Let \displaystyle z(x) = \sqrt{-x} for x \le 0 and z(x) = x^2 for x > 0. Find both one-sided limits at 0, the limit, and z(0).
Reasoning
- A function has \lim_{x \to 2^-} f(x) = 5, \lim_{x \to 2^+} f(x) = 5 and f(2) = -1. What is \lim_{x \to 2} f(x)? Explain using the criterion.
- Ingvild says: “\lim_{x \to a^-} f(x) = 3 and f(a) = 3, so \lim_{x \to a} f(x) = 3.” Give a piecewise function that shows the reasoning is wrong.
- Prove directly from the meaning of the limit that if \lim_{x \to a} f(x) = L, then \lim_{x \to a^+} f(x) = L.
- Explain why \lim_{x \to a} \lfloor x \rfloor exists exactly when a is not an integer.
- Sjurdur has two functions with f(x) = g(x) for every x > a, though f(a) \ne g(a) and the two differ for x < a. Explain why \lim_{x \to a^+} f(x) and \lim_{x \to a^+} g(x) are equal (or both fail to exist), and why this lets a piece’s rule be used at an input its condition leaves out.
Answers
- Left 18, right 25, so \lim_{w \to 30} F(w) does not exist; F(30) = 18, since 20 < w \le 30 includes 30.
- \lim_{w \to 35} F(w) = 25 and \lim_{w \to 26} F(w) = 18, inside single steps.
- \lim_{x \to 3^-} f(x) = 3^2 + 1 = 10 and \lim_{x \to 3^+} f(x) = 4 \cdot 3 - 2 = 10, so \lim_{x \to 3} f(x) = 10, and f(3) = 10.
- \lim_{x \to -2^-} g(x) = 3 - (-2) = 5 and \lim_{x \to -2^+} g(x) = (-2)^2 + 2(-2) = 0. The sides disagree, so the limit does not exist; g(-2) = 5.
- \lim_{x \to 1^-} m(x) = 1^3 = 1 and \lim_{x \to 1^+} m(x) = 2 - 1^2 = 1, so \lim_{x \to 1} m(x) = 1, while m(1) = 5.
- \lim_{x \to 4^-} p(x) = 2 \cdot 4 - 5 = 3. For x just above 4, x - 4 is small and positive, so \lim_{x \to 4^+} p(x) = \infty: no right-hand limit as a number. So \lim_{x \to 4} p(x) does not exist.
- At -1: left -1 + 4 = 3, right (-1)^2 + 2 = 3, so \lim_{x \to -1} u(x) = 3. At 0 only the middle rule is in force: \lim_{x \to 0} u(x) = 2. At 2: left 2^2 + 2 = 6, right 10 - 2 = 8, so \lim_{x \to 2} u(x) does not exist.
- \displaystyle \lim_{x \to 4^-} v(x) = \sqrt{4 + 5} = 3 and \lim_{x \to 4^+} v(x) = 4 - 1 = 3, so \lim_{x \to 4} v(x) = 3, and v(4) = 3.
- Left -2, right 2; the two-sided limit does not exist. Since |2x - 6| = 2|x - 3|, the quotient is -2 for x < 3 and 2 for x > 3.
- Left -1, right 1; no limit. The top is x(x - 1), so for x < 1 the quotient is \displaystyle \frac{x(x - 1)}{-(x - 1)} = -x, heading to -1; for x > 1 it is x, heading to 1.
- \displaystyle \lim_{x \to 4} \frac{|x - 1|}{x - 1} = 1, since near 4 every input exceeds 1; and \displaystyle \lim_{x \to -3} \frac{|x - 1|}{x - 1} = -1, since near -3 every input is below 1.
- All three are 3. For x < 3 the expression is x - (x - 3) = 3; for x > 3 it is 2x - 3, heading to 3.
- \lim_{x \to -2^-} \lfloor x \rfloor = -3 and \lim_{x \to -2^+} \lfloor x \rfloor = -2 (so no limit at -2); \lim_{x \to -1.5} \lfloor x \rfloor = -2, since \lfloor x \rfloor = -2 for -2 \le x < -1.
- Left 1, right 2, since \displaystyle \frac{x}{2} lies in [1, 2) just left of 4 and in [2, 3) just right; no limit at 4. Near 5, \displaystyle \frac{x}{2} is near 2.5, so \displaystyle \lim_{x \to 5} \left\lfloor \frac{x}{2} \right\rfloor = 2.
- Left 2 + (-3) = -1, right 3 + (-4) = -1, so \lim_{x \to 3} k(x) = -1; but k(3) = 3 + (-3) = 0.
- \lim_{w \to 40^-} F(w) = 25. The fare is defined only for 0 \le w \le 40, so there are no inputs above 40.
- \displaystyle \lim_{x \to 9^-} \sqrt{9 - x} = 0. The root needs x \le 9, so only the left-hand limit is meaningful.
- 2. As x \to 2^+, x heads to 2 and \displaystyle \sqrt{x - 2} heads to 0, so by the sum law of The limit laws the sum heads to 2 + 0 = 2.
- \displaystyle \lim_{x \to 0^-} z(x) = \sqrt{0} = 0 and \lim_{x \to 0^+} z(x) = 0^2 = 0, so \lim_{x \to 0} z(x) = 0, and z(0) = 0.
- \lim_{x \to 2} f(x) = 5. Both one-sided limits exist and equal 5, which is all the criterion asks; f(2) = -1 is never consulted.
- Take f(x) = x for x \le 3 and f(x) = x + 1 for x > 3, with a = 3. The left-hand limit and f(3) are 3, but the right-hand limit is 4, so there is no limit.
- Name a tolerance. The two-sided limit gives a distance d such that every x \ne a with |x - a| < d has |f(x) - L| below it. Every x with a < x < a + d is such an input. This holds for every tolerance, so \lim_{x \to a^+} f(x) = L.
- If a is not an integer, it lies strictly between consecutive integers n < a < n + 1, and \lfloor x \rfloor = n on that whole window, so the limit is n. If a is an integer, the left-hand limit is a - 1 and the right-hand limit is a; they differ, so there is no limit.
- The right-hand limit at a uses only inputs with a < x < a + d for small d, where f and g agree, so any closeness met by one is met by the other; values at a and to the left are never used. So a piece governing f only for x > a has the same right-hand limit as f, and for a polynomial piece that limit is its value at a.