Strategies for One-Sided Limits

Core Definition

A one-sided limit evaluates the behavior of a function as the input approaches a target value $a$ strictly from one direction along the real line:

Left-Hand Limit: $\lim_{x \to a^-} f(x)$ ( $x$ approaches $a$ with $x < a$ )
Right-Hand Limit: $\lim_{x \to a^+} f(x)$ ( $x$ approaches $a$ with $x > a$ ) One-sided limits are the fundamental building blocks for verifying the Existence Theorem for Two-Sided Limits and diagnosing Discontinuities.

Core Strategies for Evaluating One-Sided Limits

Strategy 1: The "Positive and Negative Zero" Approach ( $0^+$ vs $0^-$ )

When direct substitution into a fraction yields a non-zero numerator over zero ( $\frac{k}{0}$ where $k \neq 0$ ), the limit represents an infinite limit (vertical asymptote).

Instead of plugging in tedious decimals (like $2.999$ or $3.001$ ), use the signed zero approach:

Determine the sign of the non-zero numerator ( $k > 0$ or $k < 0$ ).
Determine whether the denominator approaches zero through positive values ( $0^+$ ) or negative values ( $0^-$ ) based on the approach direction:
- For $x \to a^+$ ( $x > a$ ), evaluate the sign of the denominator expression.
- For $x \to a^-$ ( $x < a$ ), evaluate the sign of the denominator expression.
Apply the signed-zero division rules:

Fraction Form	Denominator Sign	Output Limit
$\frac{+k}{0^+}$	Tiny positive decimal	$+\infty$
$\frac{+k}{0^-}$	Tiny negative decimal	$-\infty$
$\frac{-k}{0^+}$	Tiny positive decimal	$-\infty$
$\frac{-k}{0^-}$	Tiny negative decimal	$+\infty$

Strategy 2: Branch Selection for Piecewise Functions

For a piecewise function whose rule changes at $x = c$ :

To compute $\lim_{x \to c^-} f(x)$ , select the formula defined on the domain interval $x < c$ .
To compute $\lim_{x \to c^+} f(x)$ , select the formula defined on the domain interval $x > c$ .
Evaluate each branch independently using direct substitution or algebraic simplification.

Strategy 3: Resolving Absolute Value Quotients

To evaluate limits involving $|x - c|$ :

Rewrite the absolute value using its piecewise definition: $|x - c| = \begin{cases} +(x - c) & : x > c \\ -(x - c) & : x < c \end{cases}$
For a left-hand limit ( $x \to c^-$ ), substitute $-(x - c)$ and cancel common factors.
For a right-hand limit ( $x \to c^+$ ), substitute $+(x - c)$ and cancel common factors.

Example Problems

Example Problem 1: The Signed Zero Trick for Vertical Asymptotes

Evaluate the directional limits: (a) $\lim_{x \to 4^+} \frac{2x - 3}{x - 4}$
(b) $\lim_{x \to 4^-} \frac{2x - 3}{x - 4}$

Solution

Step 1: Check numerator value at $x = 4$
Numerator: $2(4) - 3 = 5 > 0$ (positive constant $+5$ ).

Step 2: Evaluate Part (a) ( $x \to 4^+$ )
As $x \to 4$ from the right ( $x > 4$ ), the denominator $(x - 4)$ is slightly greater than $0$ , representing a positive zero ( $0^+$ ):

\lim_{x \to 4^+} \frac{2x - 3}{x - 4} = \frac{+5}{0^+} = +\infty

Step 3: Evaluate Part (b) ( $x \to 4^-$ )
As $x \to 4$ from the left ( $x < 4$ ), the denominator $(x - 4)$ is slightly less than $0$ , representing a negative zero ( $0^-$ ):

\lim_{x \to 4^-} \frac{2x - 3}{x - 4} = \frac{+5}{0^-} = -\infty

Example Problem 2: Piecewise Function with Radical Branch

Evaluate the left- and right-sided limits at $x = 1$ for:

f(x) = \begin{cases} \sqrt{5 - x} & : x \le 1 \\ 3x^2 - 1 & : x > 1 \end{cases}

Solution

Step 1: Left-hand limit ( $x \to 1^-$ )
Use the formula for $x \le 1$ :

\lim_{x \to 1^-} \sqrt{5 - x} = \sqrt{5 - 1} = \sqrt{4} = 2

Step 2: Right-hand limit ( $x \to 1^+$ )
Use the formula for $x > 1$ :

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

Step 3: Compare results
Since $\lim_{x \to 1^-} f(x) = 2 = \lim_{x \to 1^+} f(x)$ , the two-sided limit exists and equals $2$ .

Connections & References

Parent Concepts: Limits, Calculus Foundations
Related Concepts: Existence Theorem for Two-Sided Limits, Vertical Asymptotes, Piecewise Functions, Infinite Discontinuity, Jump Discontinuity
Course Links: Calc-1 Session 02 | Calc-1 Session 03 | Calc-1 Session 04