One-Sided Limits in Detail

Understanding left-hand and right-hand limits and when they matter

One-Sided Limits Explained

Sometimes a function behaves differently depending on which direction you approach a point. One-sided limits let us describe this precisely.

Left-Hand Limit

The left-hand limit examines the function as we approach from the left (smaller values):

$\lim_{x \to a^-} f(x) = L$

Think of it as: "What happens as we walk toward a from the left side of the number line?"

Key Points:

The superscript - means "from below" or "from the left"
We only look at x-values less than a
Example: approaching x = 3 from values like 2.9, 2.99, 2.999...

Right-Hand Limit

The right-hand limit examines the function as we approach from the right (larger values):

$\lim_{x \to a^+} f(x) = L$

Think of it as: "What happens as we walk toward a from the right side of the number line?"

Key Points:

The superscript + means "from above" or "from the right"
We only look at x-values greater than a
Example: approaching x = 3 from values like 3.1, 3.01, 3.001...

The Big Rule

For a two-sided limit to exist, both one-sided limits must exist AND be equal!

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

Example: When One-Sided Limits Differ

Consider a piecewise function:

$f(x) = \begin{cases} x + 1 & \text{if } x < 2 \\ x + 5 & \text{if } x \geq 2 \end{cases}$

From the left (using $x + 1$ ): $\lim_{x \to 2^-} f(x) = 2 + 1 = 3$

From the right (using $x + 5$ ): $\lim_{x \to 2^+} f(x) = 2 + 5 = 7$

Since $3 \neq 7$ , the two-sided limit does not exist!

$\lim_{x \to 2} f(x) = \text{DNE}$

Why This Matters

One-sided limits are crucial for:

Piecewise functions that change formula at a point
Functions with jumps (discontinuities)
Rational functions near vertical asymptotes
Absolute value functions at the vertex

Visual Interpretation

On a graph:

Left-hand limit: Cover everything to the right of a, look where the curve is heading
Right-hand limit: Cover everything to the left of a, look where the curve is heading
If they point to different heights, the two-sided limit DNE

📚 Practice Problems

1Problem 1medium

❓ Question:

Find $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ where $f(x) = \begin{cases} 2x & \text{if } x < 3 \\ x^2 & \text{if } x \geq 3 \end{cases}$

💡 Show Solution

Left-hand limit (approaching from x < 3):

When x < 3, we use the formula $f(x) = 2x$

$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 2x = 2(3) = 6$

Right-hand limit (approaching from x ≥ 3):

When x ≥ 3, we use the formula $f(x) = x^2$

$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} x^2 = 3^2 = 9$

Conclusion:

Since $\lim_{x \to 3^-} f(x) = 6$ and $\lim_{x \to 3^+} f(x) = 9$ , and $6 \neq 9$ :

$\lim_{x \to 3} f(x) = \text{DNE}$

The function has a jump discontinuity at x = 3.

2Problem 2hard

❓ Question:

Given $g(x) = |x - 5|$ , find $\lim_{x \to 5^-} \frac{g(x)}{x - 5}$ and $\lim_{x \to 5^+} \frac{g(x)}{x - 5}$

💡 Show Solution

First, let's rewrite the absolute value as a piecewise function:

$|x - 5| = \begin{cases} -(x - 5) = 5 - x & \text{if } x < 5 \\ x - 5 & \text{if } x \geq 5 \end{cases}$

Left-hand limit (x < 5):

$\lim_{x \to 5^-} \frac{|x - 5|}{x - 5} = \lim_{x \to 5^-} \frac{5 - x}{x - 5} = \lim_{x \to 5^-} \frac{-(x - 5)}{x - 5} = \lim_{x \to 5^-} (-1) = -1$

Right-hand limit (x ≥ 5):

$\lim_{x \to 5^+} \frac{|x - 5|}{x - 5} = \lim_{x \to 5^+} \frac{x - 5}{x - 5} = \lim_{x \to 5^+} 1 = 1$

Result:

Left-hand limit: -1
Right-hand limit: 1
Two-sided limit: DNE (because -1 ≠ 1)

3Problem 3easy

❓ Question:

Find lim(x→3⁻) f(x) and lim(x→3⁺) f(x) where f(x) = { x + 1, if x < 3; 2x - 3, if x ≥ 3 }

💡 Show Solution

Step 1: Find left-hand limit lim(x→3⁻): Approaching 3 from left means x < 3 Use f(x) = x + 1 lim(x→3⁻) (x + 1) = 3 + 1 = 4

Step 2: Find right-hand limit lim(x→3⁺): Approaching 3 from right means x ≥ 3 Use f(x) = 2x - 3 lim(x→3⁺) (2x - 3) = 2(3) - 3 = 3

Step 3: Compare: lim(x→3⁻) f(x) = 4 lim(x→3⁺) f(x) = 3 Since 4 ≠ 3, lim(x→3) f(x) does not exist

Answer: lim(x→3⁻) f(x) = 4, lim(x→3⁺) f(x) = 3, limit DNE

4Problem 4medium

❓ Question:

Evaluate lim(x→0⁻) (|x|/x)

💡 Show Solution

Step 1: Understand the absolute value for x < 0: When x < 0, |x| = -x

Step 2: Substitute: |x|/x = (-x)/x = -1 for all x < 0

Step 3: Evaluate the limit: As x→0 from the left, f(x) = -1 constantly lim(x→0⁻) (|x|/x) = -1

Step 4: Verify with values: x = -0.1: |-0.1|/(-0.1) = 0.1/(-0.1) = -1 ✓ x = -0.01: |-0.01|/(-0.01) = 0.01/(-0.01) = -1 ✓

Answer: -1

5Problem 5hard

❓ Question:

For f(x) = 1/(x - 2), find lim(x→2⁻) f(x) and lim(x→2⁺) f(x).

💡 Show Solution

Step 1: Analyze left-hand limit (x→2⁻): As x approaches 2 from left (x < 2): • x - 2 approaches 0 from the negative side • x - 2 is small and negative • 1/(negative small number) = large negative lim(x→2⁻) 1/(x - 2) = -∞

Step 2: Create table from left: x: 1.9 1.99 1.999 x-2: -0.1 -0.01 -0.001 1/(x-2): -10 -100 -1000 → -∞

Step 3: Analyze right-hand limit (x→2⁺): As x approaches 2 from right (x > 2): • x - 2 approaches 0 from the positive side • x - 2 is small and positive • 1/(positive small number) = large positive lim(x→2⁺) 1/(x - 2) = +∞

Step 4: Create table from right: x: 2.1 2.01 2.001 x-2: 0.1 0.01 0.001 1/(x-2): 10 100 1000 → +∞

Answer: lim(x→2⁻) f(x) = -∞, lim(x→2⁺) f(x) = +∞

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