Comprehensive Study Notes on Limits: Infinite Limits, One-Sided Limits, and Asymptotic Behavior

Infinite Limits and Vertical Asymptotes

• Definition of Near $a$: A function $f(x)$ is defined near $a$ if it is defined on an open interval around $a$, such as $(a - h, a + h)$ for some positive distance $h > 0$, excluding the point $a$ itself ($x \neq a$).
• Infinite Limits: Let $f(x)$ be defined near $a$.
  • Positive Infinite Limit: $\lim_{x \to a} f(x) = \infty$ means that the values of $f(x)$ can be made arbitrarily large positive by restricting $x$ to be sufficiently close to $a$, but not equal to $a$.
  • Negative Infinite Limit: $\lim_{x \to a} f(x) = -\infty$ means that the values of $f(x)$ can be made arbitrarily large negative by restricting $x$ to be sufficiently close to $a$, but not equal to $a$.
• Vertical Asymptote and Singularity: A vertical line $x = a$ represents a vertical asymptote (or singularity) of $f(x)$ if the function blows up, approaching $\infty$ or $-\infty$ as $x$ approaches $a$.
• Three Cases of Behavior at Vertical Asymptotes:
  • Both sides approach positive infinity: $\lim_{x \to a^-} f(x) = \infty$ and $\lim_{x \to a^+} f(x) = \infty \implies \lim_{x \to a} f(x) = \infty$
  • Both sides approach negative infinity: $\lim_{x \to a^-} f(x) = -\infty$ and $\lim_{x \to a^+} f(x) = -\infty \implies \lim_{x \to a} f(x) = -\infty$
  • Opposite sides approach different infinities: One side approaches $\infty$ and the other side approaches $-\infty \implies \lim_{x \to a} f(x) = \text{DNE}$ (Does Not Exist).
• Rationale for Non-Existent Limits at Asymptotes: Limits require a single, unique value. If different paths of approach (left-hand vs. right-hand) yield conflicting infinite behavior ($\infty$ vs. $-\infty$), no unique limit value can be assigned, rendering the two-sided limit non-existent.

Behavior and Limit Evaluation of Rational Functions

• Definition of a Rational Function: A function $f(x) = \frac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are polynomials. At any given point $a$, a rational function is either defined at $a$ or has a vertical asymptote at $a$.
• Case 1: Non-Zero Denominator ($q(a) \neq 0$):
  • If $q(a) \neq 0$, then $a$ is in the domain of $f(x)$. Evaluate the limit directly using direct substitution: $\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}$
  • Example: Calculate $\lim_{x \to 3} \frac{x^2 - 1}{x^2 + x + 1}$. $\lim_{x \to 3} \frac{x^2 - 1}{x^2 + x + 1} = \frac{(3)^2 - 1}{(3)^2 + 3 + 1} = \frac{9 - 1}{9 + 3 + 1} = \frac{8}{13}$
• Case 2: Non-Zero Numerator Over Zero Denominator ($p(a) \neq 0$ and $q(a) = 0$):
  • The limit takes the form $\frac{\text{nonzero}}{0}$. This condition guarantees a vertical asymptote at $x = a$.
  • The limit must be one of three possibilities: $\infty$, $-\infty$, or $\text{DNE}$.
• Sign Analysis Principles for Products and Quotients:
  • The product or quotient of two quantities $a$ and $b$ is positive ($a \cdot b > 0$ or $\frac{a}{b} > 0$) when $a$ and $b$ have the same sign (both positive or both negative).
  • The product or quotient is negative ($a \cdot b < 0$ or $\frac{a}{b} < 0$) when $a$ and $b$ have opposite signs (one positive and one negative).
• Sign Analysis Example 1 (Positive Infinite Limit):
  • Evaluate $\lim_{x \to 1} \frac{x}{(x-1)^2}$.
  • Limit form: $\frac{1}{0}$ ($\frac{\text{nonzero}}{0}$ form).
  • Numerator sign: $x$ is positive for all $x$ near $1$
  • Denominator sign: $(x-1)^2$ is positive for all $x \neq 1$ due to the square power.
  • Ratio sign: $\frac{\text{positive}}{\text{positive}} = \text{positive}$ for all $x$ near $1$
  • Conclusion: $\lim_{x \to 1} \frac{x}{(x-1)^2} = \infty$
• Sign Analysis Example 2 (Negative Infinite Limit):
  • Evaluate $\lim_{x \to 0} \frac{1}{x^2(x-1)}$.
  • Limit form: $\frac{1}{0}$ ($\frac{\text{nonzero}}{0}$ form).
  • Denominator factoring: $x^2(x-1)$.
  • Term $x^2$: positive for all $x \neq 0$
  • Term $(x-1)$: negative for all $x$ near $0$ (since $x$ near $0$ implies $x < 1$).
  • Denominator product: $\text{positive} \times \text{negative} = \text{negative}$ for all $x$ near $0$
  • Ratio sign: $\frac{\text{positive}}{\text{negative}} = \text{negative}$ for all $x$ near $0$
  • Conclusion: $\lim_{x \to 0} \frac{1}{x^2(x-1)} = -\infty$
• Sign Analysis Example 3 (Non-Existent Limit):
  • Evaluate $\lim_{x \to 1} \frac{x}{x-1}$.
  • Limit form: $\frac{1}{0}$ ($\frac{\text{nonzero}}{0}$ form).
  • Numerator sign: $x$ is positive for all $x$ near $1$
  • Denominator sign: $(x-1)$ is positive when $x > 1$ and negative when $x < 1$
  • Right-hand approach ($x > 1$): $\frac{\text{positive}}{\text{positive}} = \text{positive} \implies \lim_{x \to 1^+} \frac{x}{x-1} = \infty$
  • Left-hand approach ($x < 1$): $\frac{\text{positive}}{\text{negative}} = \text{negative} \implies \lim_{x \to 1^-} \frac{x}{x-1} = -\infty$
  • Conclusion: Discrepancy between one-sided limits implies $\lim_{x \to 1} \frac{x}{x-1} = \text{DNE}$

Systematic Algorithm for Finding Limits of Rational Functions

• Complete Decision Algorithm for $\lim_{x \to a} \frac{p(x)}{q(x)}$:
  • Step 1: Check the denominator at $a$. If $q(a) \neq 0$, evaluate directly via substitution to find $\frac{p(a)}{q(a)}$.
  • Step 2: If $q(a) = 0$ and $p(a) \neq 0$ ($\frac{\text{nonzero}}{0}$ form), identify a vertical asymptote. Analyze the signs of the numerator and denominator near $a$ (separately for $x < a$ and $x > a$) to classify the limit as $\infty$, $-\infty$, or $\text{DNE}$.
  • Step 3: If $q(a) = 0$ and $p(a) = 0$ ($\frac{0}{0}$ indeterminate form), factor out $(x-a)$ from both $p(x)$ and $q(x)$, cancel the common factor, and repeat the decision algorithm on the simplified rational expression.
• Algebraic Simplification Example (Sum of Rational Functions):
  • Problem: Evaluate $\lim_{x \to 2} \left( \frac{1}{x-2} - \frac{4}{x^2 - 4} \right)$.
  • Apply difference of squares identity: $a^2 - b^2 = (a-b)(a+b) \implies x^2 - 4 = (x-2)(x+2)$.
  • Find a common denominator to combine fractions: $\frac{1}{x-2} - \frac{4}{(x-2)(x+2)} = \frac{x+2}{(x-2)(x+2)} - \frac{4}{(x-2)(x+2)}$
  • Combine numerators: $\frac{(x+2) - 4}{(x-2)(x+2)} = \frac{x-2}{(x-2)(x+2)}$
  • Evaluate expression form at $x = 2$: Yields $\frac{0}{0}$.
  • Apply factor and cancel step: $\frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2} \quad \text{for } x \neq 2$
  • Re-evaluate the limit using direct substitution: $\lim_{x \to 2} \frac{1}{x+2} = \frac{1}{2+2} = \frac{1}{4}$

One-Sided Limits and Existence Criteria

• Left-Handed Limit Definition: $\lim_{x \to a^-} f(x) = L$ means the values of $f(x)$ can be made arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ with $x < a$. Only values to the left of $a$ are considered.
• Right-Handed Limit Definition: $\lim_{x \to a^+} f(x) = L$ means the values of $f(x)$ can be made arbitrarily close to $L$ by restricting $x$ to be sufficiently close to $a$ with $x > a$. Only values to the right of $a$ are considered.
• Piecewise Function Evaluation Procedure:
  • Given the piecewise defined function: $f(x) = \begin{cases} x^2 - 1 & \text{if } x < 1 \\ x + 2 & \text{if } x \ge 1 \end{cases}$
  • Evaluate right-handed limit $\lim_{x \to 1^+} f(x)$: Restrict to condition $x > 1$, so $f(x) = x + 2$: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 2) = 1 + 2 = 3$
  • Evaluate left-handed limit $\lim_{x \to 1^-} f(x)$: Restrict to condition $x < 1$, so $f(x) = x^2 - 1$: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 - 1) = 1^2 - 1 = 0$
• Fundamental Existence Theorem for Two-Sided Limits:
  • Statement: A two-sided limit exists and equals $L$ if and only if both the left-handed limit and right-handed limit exist and are equal to $L$: $\lim_{x \to a} f(x) = L \iff \left( \lim_{x \to a^-} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} f(x) = L \right)$
  • Scope: The logical equivalence holds whether $L$ is a finite real number, $\infty$, or $-\infty$.
  • Application: For the piecewise function $f(x)$ above, $\lim_{x \to 1^-} f(x) = 0$ and $\lim_{x \to 1^+} f(x) = 3$. Because $0 \neq 3$, the left and right limits fail to agree, so $\lim_{x \to 1} f(x) = \text{DNE}$.
• Application of Existence Theorem to One-Sided Rational Functions:
  • Evaluate $\lim_{x \to 1} g(x)$ where: $g(x) = \begin{cases} \frac{1}{1-x} & \text{if } x < 1 \\ \frac{x}{x^2-1} & \text{if } x > 1 \end{cases}$
  • Left-handed limit ($x \to 1^-$): $x < 1 \implies 1-x > 0$. Non-zero over zero form yields $\lim_{x \to 1^-} \frac{1}{1-x} = \infty$.
  • Right-handed limit ($x \to 1^+$): $x > 1 \implies x > 0$ and $x^2-1 > 0$. Non-zero over zero form yields $\lim_{x \to 1^+} \frac{x}{x^2-1} = \infty$.
  • Conclusion: Since $\lim_{x \to 1^-} g(x) = \infty$ and $\lim_{x \to 1^+} g(x) = \infty$, the two-sided limit exists and equals $\lim_{x \to 1} g(x) = \infty$

Limits at Infinity

• Concept: Limits as $x \to \infty$ or $x \to -\infty$ specify the long-term horizontal asymptotic behavior of a function as $x$ grows arbitrarily large in the positive or negative direction.
• Power Rule Theorem for Limits at Infinity:
  • If $r > 0$ is a rational number such that $x^r$ is defined for all $x$, then: $\lim_{x \to \infty} \frac{1}{x^r} = 0$
  • If $x^r$ is defined for negative values of $x$, then: $\lim_{x \to -\infty} \frac{1}{x^r} = 0$
  • Domain Restriction Warning: The exponent $r = \frac{1}{2}$ is a non-example for $x \to -\infty$ because $x^{1/2} = \sqrt{x}$ is undefined for negative real numbers ($x < 0$).
• Strategy for Rational Functions as $x \to \pm \infty$:
  • Divide every term in both the numerator and denominator by the highest power of $x$ occurring in the denominator.
• Worked Example 1 (Finite Limit at Infinity):
  • Evaluate $\lim_{x \to \infty} \frac{x^3 - 2x + 1}{2x^3 - 8}$.
  • Identify highest denominator power: $x^3$. Divide numerator and denominator by $x^3$: $\lim_{x \to \infty} \frac{\frac{x^3}{x^3} - \frac{2x}{x^3} + \frac{1}{x^3}}{\frac{2x^3}{x^3} - \frac{8}{x^3}} = \lim_{x \to \infty} \frac{1 - \frac{2}{x^2} + \frac{1}{x^3}}{2 - \frac{8}{x^3}}$
  • Apply limit laws and power rules as $x \to \infty$: $\frac{1 - 2(0) + 0}{2 - 8(0)} = \frac{1}{2}$
• Worked Example 2 (Infinite Limit at Negative Infinity):
  • Evaluate $\lim_{x \to -\infty} \frac{x^2 - x}{1 - \frac{1}{x}}$.
  • Dividing through by highest power of $x$ in denominator: $\lim_{x \to -\infty} \frac{x \left( 1 - \frac{1}{x} \right)}{1}$
  • Substitute limit values: Numerator approaches $-\infty$, denominator equals $1$. Resulting form $\frac{-\infty}{1} = -\infty$.

Comprehensive Graphical Limit Analysis

• Graphical Evaluation Rules: For a given graph of a function $f(x)$, limits are extracted by observing function behavior along specific paths:
  • Far-Left Long-Term Behavior: $\lim_{x \to -\infty} f(x) = \infty$
  • Approach to $x = -3$ from the Left: $\lim_{x \to -3^-} f(x) = 0$
  • Two-Sided Approach to $x = -2$: Both left and right paths approach height $4$, so $\lim_{x \to -2} f(x) = 4$
  • Approach to $x = -1$ from the Right: $\lim_{x \to -1^+} f(x) = 1$
  • Approach to $x = -1$ from the Left: $\lim_{x \to -1^-} f(x) = 3$
  • Two-Sided Approach to $x = -1$: Because $\lim_{x \to -1^-} f(x) = 3 \neq 1 = \lim_{x \to -1^+} f(x)$, $\lim_{x \to -1} f(x) = \text{DNE}$
  • Approach to Vertical Asymptote at $x = 0$: Function shoots upward to infinity from both sides, so $\lim_{x \to 0} f(x) = \infty$
  • Far-Right Long-Term Behavior: Function flattens along the horizontal axis, so $\lim_{x \to \infty} f(x) = 0$