◀ Back to Note

Tony's Intertwingled Memex • Reference Manual

Existence Theorem for Two-Sided Limits

Formulas & Definitions
Core Definition

A two-sided limit lim⁡x→af(x)=L\lim_{x \to a} f(x) = Llimx→a​f(x)=L exists (where LLL is a finite real number) if and only if both the left-hand limit and the right-hand limit exist and approach the exact same value LLL:

lim⁡x→a−f(x)=Landlim⁡x→a+f(x)=L  ⟺  lim⁡x→af(x)=L\lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L \iff \lim_{x \to a} f(x) = Lx→a−lim​f(x)=Landx→a+lim​f(x)=L⟺x→alim​f(x)=L
If lim⁡x→a−f(x)≠lim⁡x→a+f(x)\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)limx→a−​f(x)=limx→a+​f(x), or if either directional limit fails to exist (blows up to ±∞\pm\infty±∞ or oscillates endlessly), the two-sided limit Does Not Exist (DNE).

Concept Explanation

The Existence Theorem establishes the principle of directional path agreement. In single-variable calculus, an input xxx can approach a target coordinate aaa from only two possible directions along the real number line:

  1. Left-Hand Limit (lim⁡x→a−f(x)\lim_{x \to a^-} f(x)limx→a−​f(x)): Evaluates the behavior of f(x)f(x)f(x) as xxx gets arbitrarily close to aaa through values strictly less than aaa (x<ax < ax<a).
  2. Right-Hand Limit (lim⁡x→a+f(x)\lim_{x \to a^+} f(x)limx→a+​f(x)): Evaluates the behavior of f(x)f(x)f(x) as xxx gets arbitrarily close to aaa through values strictly greater than aaa (x>ax > ax>a).
Left-Hand Path (x < a)           Right-Hand Path (x > a)
---------------------> [ a ] <---------------------
      x -> a^-                      x -> a^+

For the overall limit to exist, both paths must "aim at the same destination height LLL", regardless of whether the function is actually defined at x=ax = ax=a.


When Limits Fail to Exist (DNE)

A two-sided limit fails to exist in any of the following three structural scenarios:

  1. Jump Discontinuity: Both directional limits exist and are finite numbers, but they disagree (lim⁡x→a−f(x)≠lim⁡x→a+f(x)\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)limx→a−​f(x)=limx→a+​f(x)).
  2. Infinite Discontinuity (Unbounded Behavior): One or both directional limits grow infinitely large without bound (lim⁡x→a±f(x)=±∞\lim_{x \to a^\pm} f(x) = \pm\inftylimx→a±​f(x)=±∞).
  3. Wildly Oscillating Behavior: The function fluctuates infinitely fast between different values as x→ax \to ax→a, preventing convergence on any single number (e.g., sin⁡(1/x)\sin(1/x)sin(1/x) near x=0x = 0x=0).

Example Problems

Example Problem 1: Disagreeing Directional Limits (Sign Function)

Analyze the existence of lim⁡x→0∣x∣x\lim_{x \to 0} \frac{|x|}{x}limx→0​x∣x∣​.

Solution

Step 1: Evaluate the left-hand limit (x→0−x \to 0^-x→0−)
For x<0x < 0x<0, ∣x∣=−x|x| = -x∣x∣=−x. Thus:

lim⁡x→0−∣x∣x=lim⁡x→0−−xx=−1\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1x→0−lim​x∣x∣​=x→0−lim​x−x​=−1

Step 2: Evaluate the right-hand limit (x→0+x \to 0^+x→0+)
For x>0x > 0x>0, ∣x∣=x|x| = x∣x∣=x. Thus:

lim⁡x→0+∣x∣x=lim⁡x→0+xx=1\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1x→0+lim​x∣x∣​=x→0+lim​xx​=1

Step 3: Apply the Existence Theorem
Since lim⁡x→0−f(x)=−1≠1=lim⁡x→0+f(x)\lim_{x \to 0^-} f(x) = -1 \neq 1 = \lim_{x \to 0^+} f(x)limx→0−​f(x)=−1=1=limx→0+​f(x), the left and right paths do not match.

Answer: lim⁡x→0∣x∣x=DNE\lim_{x \to 0} \frac{|x|}{x} = \text{DNE}limx→0​x∣x∣​=DNE (due to a Jump Discontinuity).

Example Problem 2: Piecewise Boundary Agreement

Determine if lim⁡x→3g(x)\lim_{x \to 3} g(x)limx→3​g(x) exists for:

g(x)={2x−1:x<3x2−4:x>3g(x) = \begin{cases} 2x - 1 & : x < 3 \\ x^2 - 4 & : x > 3 \end{cases}g(x)={2x−1x2−4​:x<3:x>3​

Solution

Step 1: Left-hand limit (x→3−x \to 3^-x→3−)

lim⁡x→3−(2x−1)=2(3)−1=5\lim_{x \to 3^-} (2x - 1) = 2(3) - 1 = 5x→3−lim​(2x−1)=2(3)−1=5

Step 2: Right-hand limit (x→3+x \to 3^+x→3+)

lim⁡x→3+(x2−4)=32−4=9−4=5\lim_{x \to 3^+} (x^2 - 4) = 3^2 - 4 = 9 - 4 = 5x→3+lim​(x2−4)=32−4=9−4=5

Step 3: Apply the Existence Theorem
Because lim⁡x→3−g(x)=5=lim⁡x→3+g(x)\lim_{x \to 3^-} g(x) = 5 = \lim_{x \to 3^+} g(x)limx→3−​g(x)=5=limx→3+​g(x), the two-sided limit exists and equals 555:

lim⁡x→3g(x)=5\lim_{x \to 3} g(x) = 5x→3lim​g(x)=5
(Note: The limit exists even though g(3)g(3)g(3) is not defined in the function rule).

Example Problem 3: Enforcing Limit Existence for Control Systems

A robotics engineer models the transition speed profile v(t)v(t)v(t) of an automated arm:

v(t)={kt+2:t<4t+6:t≥4v(t) = \begin{cases} kt + 2 & : t < 4 \\ \sqrt{t} + 6 & : t \ge 4 \end{cases}v(t)={kt+2t​+6​:t<4:t≥4​
Find the parameter kkk that guarantees the speed profile is smooth and has a valid two-sided limit at t=4 st = 4\text{ s}t=4 s.

Solution

Step 1: Evaluate both one-sided limits at t=4t = 4t=4

  • Left-hand: lim⁡t→4−(kt+2)=4k+2\lim_{t \to 4^-} (kt + 2) = 4k + 2limt→4−​(kt+2)=4k+2
  • Right-hand: lim⁡t→4+(t+6)=4+6=2+6=8\lim_{t \to 4^+} (\sqrt{t} + 6) = \sqrt{4} + 6 = 2 + 6 = 8limt→4+​(t​+6)=4​+6=2+6=8

Step 2: Set directional limits equal
By the Existence Theorem, the two-sided limit exists if and only if:

4k+2=8  ⟹  4k=6  ⟹  k=32=1.54k + 2 = 8 \implies 4k = 6 \implies k = \frac{3}{2} = 1.54k+2=8⟹4k=6⟹k=23​=1.5

Answer: k=1.5k = 1.5k=1.5


Connections & References

  • Parent Concepts: Limits, Calculus Foundations
  • Related Concepts: Strategies for One-Sided Limits, Piecewise Functions, Discontinuities, Jump Discontinuity
  • Course Links: Calc-1 Session 02 | Calc-1 Session 03