One-Sided Limits

👈👉 One-Sided Limits — Notation and Idea

Part 1 of 4 — Introducing $x \to a^-$ and $x \to a^+$

Topics in This Part

🔑 Why this matters: Some functions behave very differently from the left vs. the right. One-sided limits let us describe each side separately.

💡 Why We Need One-Sided Limits

Look at $f(x) = \dfrac{|x|}{x}$.

• For $x < 0$: $|x| = -x$, so $f(x) = -1$.
• For $x > 0$: $|x| = x$, so $f(x) = +1$.

The two sides of $x = 0$ "want" different values. A two-sided limit can't capture this — we need symbols that describe each side separately.

🔑 The $-$ and $+$ Superscript Notation

⚠️ Trap: the superscript $-$ does NOT mean "negative." It means "from the left" (smaller values). $\lim_{x \to 5^-}$ uses $x = 4.9, 4.99, \ldots$ — all positive numbers.

🗣️ Reading the Symbols Aloud

Concept Check 🎯

Translate the Notation 🧮

For $f(x) = \dfrac{|x|}{x}$:

1) $\lim_{x \to 0^-} f(x) = ?$ (whole number with sign)

2) $\lim_{x \to 0^+} f(x) = ?$ (whole number with sign)

🧮 Computing One-Sided Limits

Part 2 of 4 — Piecewise functions, $|x|$, and the sign function

Topics in This Part

🔑 Why this matters: The most common one-sided-limit problems involve splitting cases — you must know which formula to use on which side.

🧩 Piecewise Functions

Suppose $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 3x - 4 & \text{if } x \ge 2 \end{cases}$

To compute $\lim_{x \to 2^-} f(x)$, use the formula for $x < 2$: $\lim_{x \to 2^-} (x^2 + 1) = 2^2 + 1 = 5.$

To compute $\lim_{x \to 2^+} f(x)$, use the formula for $x > 2$: $\lim_{x \to 2^+} (3x - 4) = 3(2) - 4 = 2.$

The two sides disagree (5 vs. 2), so $\lim_{x \to 2} f(x)$ DNE — jump.

💡 Strategy: identify which "piece" applies on each side, then substitute.

➖➕ Absolute Value Functions

Rewrite $|x - a|$ as a piecewise function: $|x - a| = \begin{cases} -(x - a) & \text{if } x < a \\ (x - a) & \text{if } x \ge a \end{cases}$

Worked example: $\displaystyle \lim_{x \to 3} \frac{|x - 3|}{x - 3}$.

• Left ($x \to 3^-$): $|x - 3| = -(x - 3)$, so the ratio is $-1$.
• Right ($x \to 3^+$): $|x - 3| = x - 3$, so the ratio is $+1$.
• Two-sided limit DNE.

🔑 Whenever you see $|x - a|$ near $x = a$, split into cases. Almost always you get $\pm 1$ on the two sides.

🔺 The Sign Function

$\operatorname{sgn}(x) = \begin{cases} -1 & x < 0 \\ 0 & x = 0 \\ +1 & x > 0 \end{cases}$

• $\lim_{x \to 0^-} \operatorname{sgn}(x) = -1$
• $\lim_{x \to 0^+} \operatorname{sgn}(x) = +1$
• Two-sided $\lim_{x \to 0} \operatorname{sgn}(x)$ DNE.

💡 The value $\operatorname{sgn}(0) = 0$ doesn't change the limit answers — limits ignore the value at the point.

Computation Check 🎯

Compute 🧮

For $g(x) = \dfrac{|x + 4|}{x + 4}$:

1) $\lim_{x \to -4^-} g(x) = ?$

2) $\lim_{x \to -4^+} g(x) = ?$

3) Does $\lim_{x \to -4} g(x)$ exist? yes or no:

✅ The Two-Sided Existence Theorem

Part 3 of 4 — When does $\lim_{x \to a} f(x)$ exist?

Topics in This Part

🔑 Why this matters: This is the defining theorem connecting one-sided and two-sided limits. AP graders look for it explicitly in justifications.

🔑 The Existence Theorem

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

Three things must all hold:

1. The left-hand limit exists.
2. The right-hand limit exists.
3. They agree on a single value $L$.

If any one fails, the two-sided limit DNE.

💡 Notice: the value $f(a)$ itself does not enter the theorem at all. The theorem is about the two arms of the function — not the dot.

📝 Worked Examples

Example 1. $f(x) = x^2$. $\lim_{x \to 3^-} f = 9$ and $\lim_{x \to 3^+} f = 9$. Both agree → $\lim_{x \to 3} f = 9$. ✅

Example 2. Piecewise $f$ with left limit 4 and right limit 4 at $x = 1$, but $f(1) = 100$. Both one-sided limits agree on 4 → $\lim_{x \to 1} f = 4$. ✅ (The value $f(1)$ doesn't change this.)

Example 3. Piecewise $f$ with left limit 2 and right limit 6 at $x = 0$. Disagreement → $\lim_{x \to 0} f$ DNE.

Example 4. $f(x) = 1/x$ at $x = 0$: left limit $= -\infty$, right limit $= +\infty$. Neither one-sided limit exists as a finite value, so the two-sided limit DNE.

🎯 Why It Matters

The theorem is the official tool for two of the most common AP tasks:

🔑 On free-response: when justifying continuity or DNE, cite the existence theorem by name or by structure.

Theorem Check 🎯

Apply the Theorem 🔽

⚠️ When Only ONE Side Exists

Part 4 of 4 — Domain restrictions and one-sided endpoints

Topics in This Part

🔑 Why this matters: AP loves to ask about $\sqrt{x}$ at $x = 0$ and similar functions where one side is undefined.

🤔 Why Only One Side Might Exist

If a function is only defined on one side of $x = a$, you can only approach $a$ from that side. The other one-sided limit doesn't exist (because the function values don't exist there).

📝 Worked Example: $\sqrt{x - 3}$ at $x = 3$

The expression $\sqrt{x - 3}$ requires $x \ge 3$.

• $\lim_{x \to 3^+} \sqrt{x - 3} = \sqrt{0} = 0$. ✅ Defined.
• $\lim_{x \to 3^-} \sqrt{x - 3}$: for $x < 3$, the radicand is negative, so $\sqrt{x - 3}$ isn't a real number. Left-hand limit doesn't exist.

💡 In AP-style problems, the answer here is "the right-hand limit equals 0; the left-hand limit does not exist."

⚖️ The Two-Sided Limit at a Domain Endpoint

What is $\lim_{x \to 3} \sqrt{x - 3}$?

By the strict existence theorem, the two-sided limit DNE because the left-hand limit doesn't exist.

⚠️ AP convention exception: for the purposes of continuity from the right at an endpoint, AP textbooks often define $\lim_{x \to 3} \sqrt{x - 3} = 0$ by treating the right-hand limit as the relevant one. Read the problem context: most AP problems will ask explicitly for $\lim_{x \to 3^+}$, sidestepping the technicality.

Endpoint Reasoning 🎯

Exit Check ✅