Existence Theorem for Two-Sided Limits

Concept Explanation

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

1. Left-Hand Limit ($\lim_{x \to a^-} f(x)$): Evaluates the behavior of $f(x)$ as $x$ gets arbitrarily close to $a$ through values strictly less than $a$ ($x < a$).
2. Right-Hand Limit ($\lim_{x \to a^+} f(x)$): Evaluates the behavior of $f(x)$ as $x$ gets arbitrarily close to $a$ through values strictly greater than $a$ ($x > a$).

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 $L$", regardless of whether the function is actually defined at $x = a$.

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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 ($\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$).
2. Infinite Discontinuity (Unbounded Behavior): One or both directional limits grow infinitely large without bound ($\lim_{x \to a^\pm} f(x) = \pm\infty$).
3. Wildly Oscillating Behavior: The function fluctuates infinitely fast between different values as $x \to a$, preventing convergence on any single number (e.g., $\sin(1/x)$ near $x = 0$).

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Example Problems

Example Problem 1: Disagreeing Directional Limits (Sign Function)

Analyze the existence of $\lim_{x \to 0} \frac{|x|}{x}$.

Solution

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

$$\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1$$

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

$$\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1$$

Step 3: Apply the Existence Theorem
Since $\lim_{x \to 0^-} f(x) = -1 \neq 1 = \lim_{x \to 0^+} f(x)$, the left and right paths do not match.

Answer: $\lim_{x \to 0} \frac{|x|}{x} = \text{DNE}$ (due to a Jump Discontinuity).

Example Problem 2: Piecewise Boundary Agreement

Determine if $\lim_{x \to 3} g(x)$ exists for:

Solution

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

$$\lim_{x \to 3^-} (2x - 1) = 2(3) - 1 = 5$$

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

$$\lim_{x \to 3^+} (x^2 - 4) = 3^2 - 4 = 9 - 4 = 5$$

Step 3: Apply the Existence Theorem
Because $\lim_{x \to 3^-} g(x) = 5 = \lim_{x \to 3^+} g(x)$, the two-sided limit exists and equals $5$:

$$\lim_{x \to 3} g(x) = 5$$

(Note: The limit exists even though $g(3)$ is not defined in the function rule).

Example Problem 3: Enforcing Limit Existence for Control Systems

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

Find the parameter $k$ that guarantees the speed profile is smooth and has a valid two-sided limit at $t = 4\text{ s}$.

Solution

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

• Left-hand: $\lim_{t \to 4^-} (kt + 2) = 4k + 2$
• Right-hand: $\lim_{t \to 4^+} (\sqrt{t} + 6) = \sqrt{4} + 6 = 2 + 6 = 8$

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

$$4k + 2 = 8 \implies 4k = 6 \implies k = \frac{3}{2} = 1.5$$

Answer: $k = 1.5$

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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