One-Sided Limits for MATH 122
Exam Relevance for MATH 122
Important for continuity and piecewise functions. Foundation for understanding limits.
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What is a One-Sided Limit?
A one-sided limit asks: what value does $f(x)$ approach as $x$ gets closer to some number from only one direction (left OR right)?
Sometimes a function behaves differently depending on which direction you approach from. Here's a perfect example:
This is the graph of $f(x) = \frac{|x|}{x}$. Notice what happens near $x = 0$:
- Coming from the left (negative x values), the function equals $-1$
- Coming from the right (positive x values), the function equals $+1$
The function "jumps" at $x = 0$ — so the limit depends on which side you approach from!
Notation
Left-Hand Limit
$$\lim_{x \to a^-} f(x)$$
The minus sign means: $x$ approaches $a$ from values less than $a$ (from the left on the number line).
Right-Hand Limit
$$\lim_{x \to a^+} f(x)$$
The plus sign means: $x$ approaches $a$ from values greater than $a$ (from the right on the number line).
The Big Rule
A two-sided limit exists if and only if BOTH one-sided limits exist AND are equal.
$$\lim_{x \to a} f(x) = L \quad \Leftrightarrow \quad \lim_{x \to a^-} f(x) = L \text{ AND } \lim_{x \to a^+} f(x) = L$$
If the left and right limits are different, the two-sided limit does not exist (DNE).
Key Variables
- $a$ — the x-value we're approaching
- $a^-$ — approaching $a$ from the left (values less than $a$)
- $a^+$ — approaching $a$ from the right (values greater than $a$)
- $L$ — the limit value (what $f(x)$ approaches)
Problem: Using the graph of $f(x) = \frac{|x|}{x}$, find the one-sided limits at $x = 0$.
Step 1: Find the left-hand limit
As $x \to 0^-$ (approaching 0 from negative values), look at the graph on the left side of 0.
The function stays at $y = -1$.
$$\lim_{x \to 0^-} \frac{|x|}{x} = -1$$
Step 2: Find the right-hand limit
As $x \to 0^+$ (approaching 0 from positive values), look at the graph on the right side of 0.
The function stays at $y = 1$.
$$\lim_{x \to 0^+} \frac{|x|}{x} = 1$$
Step 3: Does the two-sided limit exist?
Since $-1 \neq 1$, the one-sided limits are not equal.
$$\boxed{\lim_{x \to 0} \frac{|x|}{x} \text{ does not exist (DNE)}}$$
Problem: Find $\displaystyle\lim_{x \to 2} f(x)$ where:
$$f(x) = \begin{cases} x + 1 & \text{if } x < 2 \\ 5 & \text{if } x = 2 \\ x^2 & \text{if } x > 2 \end{cases}$$
Step 1: Find the left-hand limit
As $x \to 2^-$, we use the piece where $x < 2$: $f(x) = x + 1$
$$\lim_{x \to 2^-} f(x) = 2 + 1 = 3$$
Step 2: Find the right-hand limit
As $x \to 2^+$, we use the piece where $x > 2$: $f(x) = x^2$
$$\lim_{x \to 2^+} f(x) = 2^2 = 4$$
Step 3: Compare
Since $3 \neq 4$, the one-sided limits are not equal.
$$\boxed{\lim_{x \to 2} f(x) \text{ does not exist (DNE)}}$$
Note: The actual value $f(2) = 5$ doesn't matter for the limit!
Problem: Find $\displaystyle\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$
Step 1: Simplify first
$$\frac{x^2 - 1}{x - 1} = \frac{(x-1)(x+1)}{x-1} = x + 1 \quad (\text{for } x \neq 1)$$
Step 2: Check both sides
$$\lim_{x \to 1^-} (x + 1) = 2$$ $$\lim_{x \to 1^+} (x + 1) = 2$$
Both sides equal 2, so the two-sided limit exists:
$$\boxed{\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2}$$
When to Use One-Sided Limits
- Piecewise functions — different formulas on each side
- Absolute value expressions — $|x|$ changes behavior at $x = 0$
- Square roots — $\sqrt{x}$ only exists for $x \geq 0$, so you can only approach from the right at $x = 0$
- Vertical asymptotes — function may go to $+\infty$ from one side and $-\infty$ from the other
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