One-Sided Limits

A **one-sided limit** looks at a function from only one side of a point. - The **left-hand limit** $\lim_{x \to a^-} f(x) = L$ means: as $x$ gets close to $a$ through values **less than** $a$ (like $1.9, 1.99, 1.999$ for $a = 2$), $f(x)$ gets close to $L$. - The **right-hand limit** $\lim_{x \to a^+} f(x) = M$ means: as $x$ gets close to $a$ through values **greater than** $a$ (like $2.1, 2.01, 2.001$), $f(x)$ gets close to $M$. You know this notation from the lesson Limit Notation: the small sign shows the side and is not part of the number, so $3^-$ is not $-3$. In this lesson you compute these limits and use them to decide whether $\lim_{x \to a} f(x)$ exists. **The key rule** connects one-sided and two-sided limits: $$\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L$$ If the two one-sided limits are different, the two-sided limit **does not exist** (DNE). The value $f(a)$ itself never matters for any of these limits: the function may be undefined at $a$, or have a completely different value there.

Reading One-Sided Limits from a Graph

The graph of $f$ shows the line $y = x + 1$ for $x < 1$ (open circle at $(1, 2)$) and the line $y = x + 3$ for $x \ge 1$ (closed dot at $(1, 4)$). Find $\lim_{x \to 1^-} f(x)$, $\lim_{x \to 1^+} f(x)$ and $\lim_{x \to 1} f(x)$.

Follow the graph from the left: Trace the left part toward $x = 1$: the heights approach the open circle at $y = 2$ = $\lim_{x \to 1^-} f(x) = 2$

Follow the graph from the right: Trace the right part toward $x = 1$: the heights approach $y = 4$ = $\lim_{x \to 1^+} f(x) = 4$

Compare the two sides: $2 \neq 4$, so the graph jumps at $x = 1$ = The one-sided limits differ

Decide about the two-sided limit: Different one-sided limits mean no single value is approached. The closed dot $f(1) = 4$ does not change this = $\lim_{x \to 1} f(x)$ does not exist

Answer: $\lim_{x \to 1^-} f(x) = 2$, $\lim_{x \to 1^+} f(x) = 4$, so $\lim_{x \to 1} f(x)$ does not exist (even though $f(1) = 4$).

The Function |x|/x at Zero

Find $\lim_{x \to 0^-} \frac{|x|}{x}$ and $\lim_{x \to 0^+} \frac{|x|}{x}$. Does $\lim_{x \to 0} \frac{|x|}{x}$ exist?

Simplify for positive $x$: If $x > 0$, then $|x| = x$, so $\frac{|x|}{x} = \frac{x}{x} = 1$ = $\lim_{x \to 0^+} \frac{|x|}{x} = 1$

Simplify for negative $x$: If $x < 0$, then $|x| = -x$, so $\frac{|x|}{x} = \frac{-x}{x} = -1$ = $\lim_{x \to 0^-} \frac{|x|}{x} = -1$

Check with numbers: $\frac{|0.01|}{0.01} = 1$ and $\frac{|-0.01|}{-0.01} = -1$ = The table agrees

Compare: $-1 \neq 1$ = The two-sided limit does not exist

Answer: The left-hand limit is $-1$ and the right-hand limit is $1$, so $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. The function is not even defined at $x = 0$, but that is not the reason: the reason is that the two sides disagree.

A Piecewise Function Whose Limit Exists

Let $g(x) = x^2$ for $x < 2$, $g(2) = 7$, and $g(x) = 3x - 2$ for $x > 2$. Find $\lim_{x \to 2} g(x)$.

Left side: use the piece for $x < 2$: $x^2 \to 2^2 = 4$ = $\lim_{x \to 2^-} g(x) = 4$

Right side: use the piece for $x > 2$: $3x - 2 \to 3(2) - 2 = 4$ = $\lim_{x \to 2^+} g(x) = 4$

Compare the one-sided limits: Both sides approach $4$ = $\lim_{x \to 2} g(x) = 4$

Look at the value $g(2)$: $g(2) = 7$, but the limit only cares about values near $2$ = Limit $4$, value $7$

Answer: $\lim_{x \to 2} g(x) = 4$. The isolated value $g(2) = 7$ does not change the limit; it only creates a hole with a separate dot above it.

An Absolute Value in the Denominator

Evaluate $\lim_{x \to 2^-} \frac{x^2 - 4}{|x - 2|}$ and $\lim_{x \to 2^+} \frac{x^2 - 4}{|x - 2|}$.

Factor the numerator: $x^2 - 4 = (x - 2)(x + 2)$ = $\frac{(x - 2)(x + 2)}{|x - 2|}$

Right side: $x > 2$: $|x - 2| = x - 2$, so the fraction is $x + 2 \to 4$ = $\lim_{x \to 2^+} \frac{x^2 - 4}{|x - 2|} = 4$

Left side: $x < 2$: $|x - 2| = -(x - 2)$, so the fraction is $-(x + 2) \to -4$ = $\lim_{x \to 2^-} \frac{x^2 - 4}{|x - 2|} = -4$

Check with a table: $x = 2.01$: $\frac{0.0401}{0.01} = 4.01$; $x = 1.99$: $\frac{-0.0399}{0.01} = -3.99$ = Values near $4$ and near $-4$

Answer: $\lim_{x \to 2^-} \frac{x^2 - 4}{|x - 2|} = -4$ and $\lim_{x \to 2^+} \frac{x^2 - 4}{|x - 2|} = 4$, so the two-sided limit at $2$ does not exist.

Choosing a Constant So the Limit Exists

Let $h(x) = 2x + k$ for $x < 3$ and $h(x) = x^2 - 1$ for $x \ge 3$. Find $k$ so that $\lim_{x \to 3} h(x)$ exists.

Right-hand limit: $x^2 - 1 \to 3^2 - 1 = 8$ = $\lim_{x \to 3^+} h(x) = 8$

Left-hand limit (with $k$): $2x + k \to 2(3) + k = 6 + k$ = $\lim_{x \to 3^-} h(x) = 6 + k$

Set the one-sided limits equal: $6 + k = 8$ = $k = 2$

Check: With $k = 2$: left $2(3) + 2 = 8$, right $8$ = $\lim_{x \to 3} h(x) = 8$

Answer: $k = 2$. Then both one-sided limits equal $8$, so $\lim_{x \to 3} h(x) = 8$.

One-sided limits are the tool for functions that change their rule at a point: - **Piecewise functions**: tax brackets, phone plans and shipping prices switch formulas at certain values, and you must look at each side separately - **Jumps**: a parking fee that jumps from 3 to 5 dollars after one hour has different left-hand and right-hand limits - **Absolute values**: expressions like $\frac{|x|}{x}$ behave differently for negative and positive $x$ - **Continuity**: in the lessons Continuity and Types of Discontinuity you will check one-sided limits to classify every break in a graph - **Calculus**: derivatives of functions like $|x|$ are studied with left and right limits Whenever a graph has a break, the first question is always: what happens on the left, and what happens on the right?

Mistake: Using the closed dot $f(a)$ as the limit

Why: A limit describes the heights the graph approaches near $a$, not the height at $a$. In Example 1, $f(1) = 4$, but from the left the graph approaches $2$.

Correct: Trace each side of the graph toward $a$ and ignore the dot at $x = a$. Use $f(a)$ only when the question asks for the value.

Mistake: Reading $x \to 3^-$ as "$x$ approaches $-3$"

Why: The small minus sign is written after the number, as a superscript. It describes the direction, not the sign of the number.

Correct: $x \to 3^-$ means $x = 2.9, 2.99, 2.999, \dots$ (values below $3$). $x \to 3^+$ means $x = 3.1, 3.01, 3.001, \dots$

Mistake: Using the piece that contains $x = a$ for both sides

Why: In $f(x) = x + 1$ for $x < 1$ and $f(x) = x + 3$ for $x \ge 1$, the piece $x + 3$ only describes the right side. The left side is always given by the piece for $x < 1$.

Correct: For $x \to a^-$ use the formula for values just below $a$; for $x \to a^+$ use the formula for values just above $a$. The $\le$ or $\ge$ sign only decides $f(a)$.

Mistake: Saying the limit exists because both one-sided limits exist

Why: Both one-sided limits in Example 1 exist ($2$ and $4$), but they are different numbers, so there is no single value $L$.

Correct: The two-sided limit exists only if both one-sided limits exist **and** are equal.

Parking Fees

Many parking garages charge by started hour, so the price jumps at each full hour. One-sided limits describe the price just before and just after the jump.

A garage charges $P(t) = 3$ dollars for $0 < t \le 1$ hour and $P(t) = 5$ dollars for $1 < t \le 2$ hours. Just before one hour you pay 3 dollars, just after it 5 dollars.

Building a Smooth Ramp

Engineers who join a ramp to a platform need the heights from both sides to meet. Otherwise there is a step that people can trip over.

A ramp has height $h(x) = 0.2x$ meters for $0 \le x < 5$, and the platform starts at $x = 5$ at a constant height $c$. The ramp and the platform meet without a step exactly when $\lim_{x \to 5^-} h(x) = c$.

Mobile Data Plans

A data plan often has a fixed price up to a limit and an extra charge per gigabyte after it. One-sided limits show whether the price jumps at the limit.

A plan costs $C(g) = 10$ dollars for $0 \le g \le 5$ GB and $C(g) = 10 + 2(g - 5)$ dollars for $g > 5$. Just above 5 GB the price starts to grow, but it starts from 10.

$\lim_{x \to a^-} f(x)$ is the left-hand limit: $x$ approaches $a$ through values less than $a$

$\lim_{x \to a^+} f(x)$ is the right-hand limit: $x$ approaches $a$ through values greater than $a$

$\lim_{x \to a} f(x) = L$ exactly when both one-sided limits exist and both equal $L$

If the one-sided limits are different (a jump), the two-sided limit does not exist

For a piecewise function, use the piece for values just below $a$ on the left and the piece for values just above $a$ on the right

The value $f(a)$ (open circle or closed dot) never changes a limit

Q: Can a one-sided limit exist when the two-sided limit does not?

A: Yes, and this is the typical case at a jump. For $\frac{|x|}{x}$ at $0$ both one-sided limits exist ($-1$ and $1$), but the two-sided limit does not, because they are different.

Q: Does the closed dot or the open circle decide the limit?

A: Neither. Circles and dots only show the value $f(a)$. The limits come from the curve on each side as it approaches $x = a$.

Q: What if the function is only defined on one side?

A: Then only one one-sided limit makes sense. For $\sqrt{x}$ at $0$ there are no values to the left, so we use $\lim_{x \to 0^+} \sqrt{x} = 0$. At the end of a domain, the one-sided limit is the natural question.

Q: Can a one-sided limit be infinite?

A: Yes. For example $\frac{1}{x}$ grows without bound as $x \to 0^+$. These cases are covered in the lesson Infinite Limits; in this lesson all one-sided limits are numbers.

Exit Presentation

One-Sided Limits

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Pre-Calculus / Limits

One-Sided Limits

Learn to find left-hand and right-hand limits from graphs and piecewise formulas, and decide when a two-sided limit exists.

Advanced30 min

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