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โˆฌDifferential Calculus Unit 2 Review

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2.2 One-sided limits and limit laws

โˆฌDifferential Calculus
Unit 2 Review

2.2 One-sided limits and limit laws

Written by the Fiveable Content Team โ€ข Last updated September 2025
Written by the Fiveable Content Team โ€ข Last updated September 2025
โˆฌDifferential Calculus
Unit & Topic Study Guides

Limits help us understand how functions behave near specific points. One-sided limits focus on approaching from the left or right, while limit laws and theorems provide tools for evaluating complex limits.

These concepts are crucial for grasping function behavior and continuity. By mastering limits, you'll be better equipped to tackle more advanced calculus topics like derivatives and integrals.

One-Sided Limits

Left-hand vs right-hand limits

  • Left-hand limit $\lim_{x \to a^-} f(x)$ considers function values as $x$ approaches $a$ from the left (smaller values)
  • Right-hand limit $\lim_{x \to a^+} f(x)$ considers function values as $x$ approaches $a$ from the right (larger values)
  • Limit exists only if both left-hand and right-hand limits exist and are equal ($\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$)
    • Example: $f(x) = |x|$ at $x = 0$, left-hand limit is 0, right-hand limit is 0, so the limit exists and equals 0
    • Example: $f(x) = \frac{1}{x}$ at $x = 0$, left-hand limit is $-\infty$, right-hand limit is $+\infty$, so the limit does not exist

Evaluation of one-sided limits

  • Tables: Create a table with $x$ values approaching the point from left and right, observe function value trends
    • Example: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$, table shows left-hand and right-hand limits both approach 2
  • Graphs: Visually inspect the graph near the point, determine $y$-values the function approaches from left and right
    • Example: $f(x) = \begin{cases} x^2 & x < 0 \ x & x \geq 0 \end{cases}$ at $x = 0$, graph shows left-hand limit is 0 and right-hand limit is 0
  • Algebraic manipulation: Simplify the function, factor and cancel terms, substitute the point to evaluate the limit
    • Example: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} (x + 2) = 4$, factoring and canceling $(x - 2)$ allows direct substitution

Limit Laws and Theorems

Limit laws for function operations

  • Sum rule: $\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$
  • Difference rule: $\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)$
  • Product rule: $\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$
  • Quotient rule: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}$, provided $\lim_{x \to a} g(x) \neq 0$
  • Constant multiple rule: $\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x)$, where $c$ is a constant
  • Power rule: $\lim_{x \to a} [f(x)]^n = [\lim_{x \to a} f(x)]^n$, where $n$ is a constant
    • Example: $\lim_{x \to 2} (3x^2 + 2x - 1) = 3 \cdot \lim_{x \to 2} x^2 + 2 \cdot \lim_{x \to 2} x - \lim_{x \to 2} 1 = 3 \cdot 4 + 2 \cdot 2 - 1 = 15$

Theorems for limit evaluation

  • Squeeze Theorem (Sandwich Theorem): If $f(x) \leq g(x) \leq h(x)$ near $a$ (except possibly at $a$) and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$
    • Useful for finding limits of functions "squeezed" between two other functions with known limits
    • Example: $0 \leq \frac{\sin x}{x} \leq 1$ for $x > 0$, and $\lim_{x \to 0^+} 0 = \lim_{x \to 0^+} 1 = 0$, so $\lim_{x \to 0^+} \frac{\sin x}{x} = 0$
  • Limit Comparison Test: Compares the behavior of two functions near a point
    1. If $\lim_{x \to a} \frac{f(x)}{g(x)} = L$, where $L$ is a finite positive number, then $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ either both exist or both diverge
    2. If $\lim_{x \to a} \frac{f(x)}{g(x)} = 0$ and $\lim_{x \to a} g(x)$ exists, then $\lim_{x \to a} f(x) = 0$
    • Example: To find $\lim_{x \to \infty} \frac{2x + 1}{3x - 2}$, compare with $\lim_{x \to \infty} \frac{1}{x}$. Since $\lim_{x \to \infty} \frac{\frac{2x + 1}{3x - 2}}{\frac{1}{x}} = \frac{2}{3}$, both limits exist, and $\lim_{x \to \infty} \frac{2x + 1}{3x - 2} = \frac{2}{3}$