Computing One-Sided Limits

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Textbook Reference

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Key idea

You already know what a two-sided limit means: $f(x)$ approaches $L$ from both sides as $x \to a$. A one-sided limit asks about only one side -- the left ($x \to a^-$) or the right ($x \to a^+$).

The key theorem connects them: the two-sided limit exists if and only if both one-sided limits exist and agree. When they disagree, the two-sided limit does not exist -- but each one-sided limit may still be a well-defined number that is useful on its own.

One-sided limits appear constantly with piecewise functions (where different formulas apply on different sides) and at boundaries like $x = 0$ for $\sqrt{x}$ or at the endpoints of a closed interval.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• State the definition of $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$
• Evaluate $f(x)$ for a piecewise-defined function at a given $x$
• Apply direct substitution within each piece of a piecewise function
• State the criterion: $\lim_{x \to a} f(x) = L$ if and only if both one-sided limits equal $L$

Quick Reference

One-sided limit notation.

• $\lim_{x \to a^-} f(x) = L$: $f(x)$ approaches $L$ as $x$ approaches $a$ from the left ($x < a$).
• $\lim_{x \to a^+} f(x) = L$: $f(x)$ approaches $L$ as $x$ approaches $a$ from the right ($x > a$).

Two-sided limit theorem. \[ \lim_{x \to a} f(x) = L \quad \Longleftrightarrow \quad \lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L. \]

Evaluation method for piecewise functions. For $\lim_{x \to a^-}$, use the formula that applies when $x < a$. For $\lim_{x \to a^+}$, use the formula that applies when $x > a$.

Key Concepts

1. Computing One-Sided Limits from Piecewise Formulas

Given a piecewise function, approach $a$ from the appropriate side and apply direct substitution to the formula valid on that side.

Example 1. Let \[ f(x) = \begin{cases} x^2 + 1 & x < 2 \\ 3x - 1 & x \geq 2 \end{cases}. \] Evaluate $\lim_{x \to 2^-} f(x)$ and $\lim_{x \to 2^+} f(x)$.

For $x \to 2^-$: use $x^2 + 1$ (valid when $x < 2$): \[ \lim_{x \to 2^-} (x^2 + 1) = 4 + 1 = 5. \]

For $x \to 2^+$: use $3x - 1$ (valid when $x \geq 2$): \[ \lim_{x \to 2^+} (3x - 1) = 6 - 1 = 5. \]

Both one-sided limits equal 5, so $\lim_{x \to 2} f(x) = 5$.

Example 2. Let \[ g(x) = \begin{cases} x + 3 & x < 1 \\ x^2 & x > 1 \\ 7 & x = 1 \end{cases}. \] Find $\lim_{x \to 1^-} g(x)$, $\lim_{x \to 1^+} g(x)$, and $\lim_{x \to 1} g(x)$.

$\lim_{x \to 1^-} g(x) = 1 + 3 = 4$.

$\lim_{x \to 1^+} g(x) = 1^2 = 1$.

Since $4 \neq 1$, the two-sided limit $\lim_{x \to 1} g(x)$ does not exist.

The value $g(1) = 7$ is irrelevant to the limit. The limit asks about nearby values of $x$, not the value at $x = 1$ itself.

2. One-Sided Limits and Absolute Value

Absolute value creates piecewise behavior: $|x - a| = x - a$ when $x > a$ and $|x - a| = -(x-a)$ when $x < a$. Limits involving absolute value often require one-sided treatment.

Example 3. Evaluate $\displaystyle\lim_{x \to 0^+} \frac{|x|}{x}$ and $\displaystyle\lim_{x \to 0^-} \frac{|x|}{x}$.

For $x > 0$: $|x| = x$, so $\dfrac{|x|}{x} = 1$. Right limit $= 1$.

For $x < 0$: $|x| = -x$, so $\dfrac{|x|}{x} = -1$. Left limit $= -1$.

Since $1 \neq -1$, $\lim_{x \to 0} \dfrac{|x|}{x}$ does not exist.

Example 4. Evaluate $\displaystyle\lim_{x \to 3} \frac{|x - 3|}{x - 3}$.

For $x > 3$: $|x-3| = x - 3$, so the expression equals $1$. Right limit $= 1$.

For $x < 3$: $|x-3| = -(x-3)$, so the expression equals $-1$. Left limit $= -1$.

Two-sided limit does not exist.

3. One-Sided Limits at Restricted Domains

At a left endpoint of a domain (such as $x = 0$ for $\sqrt{x}$), only the right-hand limit makes sense: \[ \lim_{x \to 0^+} \sqrt{x} = 0. \] The function is continuous at $x = 0$ from the right.

Example 5. Evaluate $\displaystyle\lim_{x \to 0^+} \frac{\sqrt{x}}{x}$.

For $x > 0$: $\dfrac{\sqrt{x}}{x} = \dfrac{1}{\sqrt{x}}$, which grows without bound as $x \to 0^+$.

The right-hand limit is $+\infty$. There is a vertical asymptote at $x = 0$.

4. The Two-Sided Criterion in Practice

To determine whether a two-sided limit exists:

1. Compute $\lim_{x \to a^-} f(x)$ using the formula valid for $x < a$.
2. Compute $\lim_{x \to a^+} f(x)$ using the formula valid for $x > a$.
3. If both are finite and equal, the two-sided limit equals that common value.
4. If they differ, or if one or both are infinite or undefined, the two-sided limit does not exist.

Common Errors

Leveled Practice

Level 1 -- Piecewise Functions

Problem 1. Let $f(x) = \begin{cases} 2x + 1 & x \leq 3 \\ x^2 - 4 & x > 3 \end{cases}$. Find $\lim_{x \to 3^-} f(x)$, $\lim_{x \to 3^+} f(x)$, and $\lim_{x \to 3} f(x)$ if it exists.

Problem 2. Let $g(x) = \begin{cases} x^2 - 2 & x < 1 \\ 3x - 4 & x \geq 1 \end{cases}$. Find $\lim_{x \to 1} g(x)$.

Problem 3. Let $h(x) = \begin{cases} x^2 & x < 0 \\ x + 1 & 0 \leq x < 2 \\ 5 - x & x \geq 2 \end{cases}$. Evaluate $\lim_{x \to 0} h(x)$ and $\lim_{x \to 2} h(x)$.

Level 2 -- Absolute Value and Parameters

Problem 4. Evaluate $\displaystyle\lim_{x \to -2} \frac{|x + 2|}{x + 2}$.

Problem 5. Find $c$ so that $\lim_{x \to 2} f(x)$ exists, where $f(x) = \begin{cases} cx^2 - 1 & x < 2 \\ x + 5 & x \geq 2 \end{cases}$.

Level 3 -- Analysis

Problem 6. Let $\lfloor x \rfloor$ denote the floor function (greatest integer $\leq x$). Find $\lim_{x \to 2^-} \lfloor x \rfloor$ and $\lim_{x \to 2^+} \lfloor x \rfloor$.

Problem 7. Prove that $\displaystyle\lim_{x \to 0} x \sin\!\left(\frac{1}{x}\right) = 0$ using the following one-sided limit approach: argue that $-|x| \leq x\sin(1/x) \leq |x|$ for $x \neq 0$, and apply the squeeze theorem on each side.

Common Misconceptions

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• Evaluate both one-sided limits of a piecewise function at a boundary point
• Write $|x - a|$ as a piecewise expression and use it in limit calculations
• Apply the two-sided limit criterion to determine whether $\lim_{x \to a} f(x)$ exists
• Find parameter values that make a two-sided limit exist
• Identify when only a one-sided limit makes sense (restricted domain)
• Distinguish clearly between $\lim_{x \to a} f(x)$ and $f(a)$

Mental Model

Think of a one-sided limit as asking: “If I walk along the graph toward $a$ from the left (or right), what height am I heading toward?”

The two-sided limit exists when both walkers -- one from each side -- arrive at the same height. A jump discontinuity is when the left-approaching walker arrives at a different height than the right-approaching walker. Neither is wrong; the function simply does not connect at $a$.

Piecewise functions make this explicit: different formulas govern each direction, so the limits can naturally differ at a boundary.

Connections

Within Calculus I (MATH161)

• Continuity: A function is continuous at $a$ if both one-sided limits equal $f(a)$. One-sided limits are the computational tool for testing continuity at piecewise boundaries.
• Infinite limits: When a one-sided limit is $+\infty$ or $-\infty$, the function has a vertical asymptote from that side.
• Differentiability: The left and right derivatives at a corner are one-sided limits of the difference quotient. If they differ, the function is not differentiable there, even if it is continuous.

Back to Calculus I Skills | Previous: Algebraic Techniques for Limits | Next: Squeeze Theorem