Precise Definition: One-Sided Limits

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Textbook Reference

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Try This First: Same Game, One Direction

For $\lim_{x \to 0^+} \sqrt{x} = 0$: the square root is only defined for $x \geq 0$, so only the right-hand limit makes sense at $x = 0$.

The challenge game: opponent names $\varepsilon > 0$, you name $\delta > 0$ so that $|\sqrt{x} - 0| < \varepsilon$ whenever $0 < x < \delta$.

Predict: $|\sqrt{x}| < \varepsilon$ when $x < \varepsilon^2$. So choose $\delta = \varepsilon^2$.

Is this right? Verify: if $0 < x < \varepsilon^2$, then $\sqrt{x} < \varepsilon$. Yes -- because $\sqrt{\cdot}$ is increasing.

Quantity-First Framing

One-sided limits ask about approach from only one direction. The precise definitions are small modifications of the two-sided definition: instead of requiring $0 < |x - a| < \delta$ (both sides), you require only the approach from the left ($a - \delta < x < a$) or from the right ($a < x < a + \delta$).

Prerequisite Check

• State the two-sided epsilon-delta definition
• State that the two-sided limit exists if and only if both one-sided limits are equal
• Compute $\lim_{x \to 0^+} \sqrt{x}$ numerically or from continuity

Quick Reference

Right-hand limit. $\lim_{x \to a^+} f(x) = L$ means: for every $\varepsilon > 0$, there exists $\delta > 0$ such that if $a < x < a + \delta$, then $|f(x) - L| < \varepsilon$.

Left-hand limit. $\lim_{x \to a^-} f(x) = L$ means: for every $\varepsilon > 0$, there exists $\delta > 0$ such that if $a - \delta < x < a$, then $|f(x) - L| < \varepsilon$.

Key difference from two-sided. Replace $0 < |x - a| < \delta$ with the appropriate one-sided interval. The output condition $|f(x) - L| < \varepsilon$ is unchanged.

Key Concepts

1. The Right-Hand Limit of $\sqrt{x}$ at 0

Example 1. Prove $\lim_{x \to 0^+} \sqrt{x} = 0$.

Scratch. Need $|\sqrt{x} - 0| < \varepsilon$, i.e., $\sqrt{x} < \varepsilon$, i.e., $x < \varepsilon^2$. For $x > 0$: choose $\delta = \varepsilon^2$.

Formal proof.

Given $\varepsilon > 0$. Let $\delta = \varepsilon^2$.

Suppose $0 < x < \delta = \varepsilon^2$.

Then $\sqrt{x} < \sqrt{\varepsilon^2} = \varepsilon$ (since $\sqrt{\cdot}$ is increasing on $[0, \infty)$).

So $|\sqrt{x} - 0| = \sqrt{x} < \varepsilon$. $\square$

2. Left-Hand Limit: Piecewise Functions

Example 2. Let $f(x) = \begin{cases} 2x + 1 & x < 3 \\ 7 & x = 3 \\ x^2 & x > 3 \end{cases}$.

Prove $\lim_{x \to 3^-} f(x) = 7$.

For $x < 3$: $f(x) = 2x + 1$.

$L = 2(3) + 1 = 7$.

$|f(x) - 7| = |(2x+1) - 7| = |2x - 6| = 2|x - 3|$.

For $x < 3$: $|x - 3| = 3 - x$, so $|f(x) - 7| = 2(3 - x) = 2|x - 3|$.

Choose $\delta = \varepsilon/2$. If $3 - \delta < x < 3$: $|x - 3| < \delta$, so $|f(x) - 7| = 2|x-3| < 2\delta = \varepsilon$. $\square$

3. Two Representations

Number line. For the right-hand limit at $a = 0$: the interval $0 < x < \delta$ is the open interval just to the right of 0. For the left-hand limit at $a = 3$: the interval $3 - \delta < x < 3$ is the open interval just to the left of 3.

Table. For $\lim_{x \to 0^+} \sqrt{x} = 0$ with $\delta = \varepsilon^2$:

4. Connecting One-Sided and Two-Sided

Theorem. $\lim_{x \to a} f(x) = L$ if and only if $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = L$.

This is a rigorous version of the informal criterion: the two-sided limit exists iff both one-sided limits agree.

The epsilon-delta proof: suppose both one-sided limits equal $L$ with respective $\delta_L$ and $\delta_R$ for a given $\varepsilon$. Take $\delta = \min(\delta_L, \delta_R)$: for $0 < |x - a| < \delta$, either $x < a$ (use $\delta_L$) or $x > a$ (use $\delta_R$); both cases give $|f(x) - L| < \varepsilon$.

5. Ask Why: Why Is the One-Sided Definition Just a Modification?

The two-sided limit allows $x$ to approach from either side: the condition $0 < |x - a| < \delta$ includes both $a - \delta < x < a$ and $a < x < a + \delta$. The one-sided definitions simply restrict to one half.

Every epsilon-delta technique carries over: finding $\delta$ in scratch work, using $\min$, bounding a quadratic factor. The only change is the allowed direction of approach.

Named Misconception: limit-equals-function-value

In Example 2, $f(3) = 7$. Is it a coincidence that $\lim_{x \to 3^-} f(x) = 7$ equals $f(3) = 7$?

Yes and no. The left-hand limit equals 7 because the formula $2x + 1$ for $x < 3$ gives $2(3)+1 = 7$ in the limit -- this is about the approach, not the value. The fact that $f(3) = 7$ is separately assigned in the piecewise definition.

The misconception: assuming the limit must equal the function value. Here they happen to agree; in general, a one-sided limit and the function value can differ. The limit is about $x$ near $a$ from one side; the function value is about $x = a$ itself.

Common Errors

Leveled Practice

Level 1 -- Right-Hand Limits

Problem 1. Prove $\lim_{x \to 1^+} (2x - 1) = 1$.

Problem 2. Prove $\lim_{x \to 0^+} x^2 = 0$.

Level 2 -- Left-Hand Limits

Problem 3. Prove $\lim_{x \to 2^-} (x + 3) = 5$.

Level 3 -- Low-Floor-High-Ceiling Extension

Problem 4 (Extension).

(a) (Floor) For $f(x) = |x|/x$: prove $\lim_{x \to 0^+} f(x) = 1$ and $\lim_{x \to 0^-} f(x) = -1$.

(b) (Mid) Use your answers from (a) and the two-sided criterion to prove that $\lim_{x \to 0} |x|/x$ does not exist.

(c) (Ceiling) For the piecewise function in Example 2: prove $\lim_{x \to 3^+} f(x) = 9$ and conclude that $\lim_{x \to 3} f(x)$ does not exist.

Common Misconceptions

Mastery Checklist

• State the epsilon-delta definitions of the left-hand and right-hand limits
• Identify the one-sided interval conditions ($a < x < a + \delta$ vs. $a - \delta < x < a$)
• Prove a right-hand or left-hand limit for a polynomial or square-root function
• Use the one-sided-to-two-sided criterion to determine when a two-sided limit fails to exist

Mental Model

The one-sided limit definitions are the two-sided definition with one half of the input interval removed. Everything else -- the challenge game, the output condition $|f(x) - L| < \varepsilon$, the scratch work, the formal proof -- is identical.

One-sided limits are essential wherever a function is only defined on one side of $a$ (like $\sqrt{x}$ at 0) or where left and right behaviors differ (like a piecewise function at its boundary). The two-sided limit is then defined through both one-sided limits agreeing.

Connections

Within MATH161

• Continuity from one side: A function is right-continuous at $a$ if $\lim_{x \to a^+} f(x) = f(a)$; left-continuous if $\lim_{x \to a^-} f(x) = f(a)$. The one-sided limit definitions make these precise.
• Intermediate Value Theorem: The IVT proof uses continuity on a closed interval, which involves one-sided continuity at the endpoints.

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