Limit Proofs and Limit Laws for Functions

Conclusion of Limit Proof using Double Implications

• The validity of the proof relies on all intermediate statements being double implications (if and only if statements, $A ext{ implies } B$ and $B ext{ implies } A$).
• Therefore, if 0 < |x-1| < rac{ ext{epsilon}}{3} , it implies that |f(x)-2| < ext{epsilon} .
• This directly matches the definition of a limit, confirming that $ext{lim}_{x \to 1} f(x) = 2$.

Clarification on $|x| > 0$ in Limits

• Assumption: The absolute value of any real number is always greater than or equal to zero (e.g., |x| ootnote{e.g., if we were talking about x-1 $as in the previous example, then$ |x-1| ootnote{It implicitly handles the requirement for |x-1| > 0 because the original limit statement itself assumes $x \neq 1$ by its very definition; the limit is concerned with values near $x=1$, but not at $x=1$.}

Handling Negative Constants in Absolute Values

• When pulling a constant out of an absolute value, such as $|-3|$, the absolute value is first evaluated to a positive number ($3$).
• Thus, if a factor like $-3$ is inside an absolute value, it would become $3$ when factored out (e.g., $|-3x| = 3|x|$).

Transition to Limit Laws for Computing Limits

• While understanding and using the formal definition of a limit for proofs is essential, performing such a rigorous proof for every limit evaluation is impractical.
• Limit Laws are a set of rules derived directly from the definition of the limit.
• Once a limit law is proven using the definition, it can be applied directly to evaluate limits of functions without re-proving the underlying principle each time.

Conditions for Limit Laws

• Let $ext{lim}<em>{x \to a} f(x) = L$ and $ext{lim}</em>{x \to a} g(x) = M$, where $L$ and $M$ are finite numbers (i.e., the limits exist).
• Let $c$ be a real number constant.

Basic Limit Laws

1. Limit of a Constant Function: $ext{lim}_{x \to a} c = c$
   • As $x$ approaches $a$, a function that is always equal to $c$ will approach $c$.
2. Limit of the Identity Function: $ext{lim}_{x \to a} x = a$
   • As $x$ approaches $a$, the value of $x$ itself approaches $a$.

Combination Limit Laws

1. Sum/Difference Law: $ext{lim}_{x \to a} [f(x) \pm g(x)] = L \pm M$
   • The limit of a sum or difference of functions is the sum or difference of their individual limits.
2. Constant Multiple Law: $ext{lim}_{x \to a} [c \cdot f(x)] = c \cdot L$
   • The limit of a constant times a function is the constant times the limit of the function.
3. Product Law: $ext{lim}_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
   • The limit of a product of functions is the product of their individual limits.
4. Quotient Law: $ext{lim}_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$, provided that $M \neq 0$.
   • The limit of a quotient of functions is the quotient of their individual limits, as long as the denominator's limit is not zero.
5. Composition (or Power) Law: $ext{lim}<em>{x \to a} f(g(x)) = f( ext{lim}</em>{x \to a} g(x)) = f(M)$, provided that $f$ is a continuous function.
   • For now, assume most basic functions (polynomials, roots, trig functions) are continuous on their domains.

Important Notes on Limit Laws

• Proof Origin: Limit laws are derived and proven using the definition of the limit ($ext{epsilon-delta}$ definition).
• Existence Requirement: Limit laws are only applicable if both individual limits ($L$ and $M$) exist (i.e., are finite numbers).

Example 1: Direct Application of Limit Laws

• Problem: Evaluate $ext{lim}_{x \to 1} (3x^3 - x^2 + 4)$.
• Process: Apply the sum/difference, constant multiple, product, and basic limit laws.
  • $ext{lim}<em>{x \to 1} 3x^3 = 3 \cdot ( ext{lim}</em>{x \to 1} x)^3 = 3 \cdot (1)^3 = 3$.
  • $ext{lim}<em>{x \to 1} x^2 = ( ext{lim}</em>{x \to 1} x)^2 = (1)^2 = 1$.
  • $ext{lim}_{x \to 1} 4 = 4$.
• Result: $3 - 1 + 4 = 6$.

Example 2: Indeterminate Form and the Conjugate Method

• Problem: Evaluate $ext{lim}_{x \to 7} \frac{\sqrt{x+2} - 3}{x-7}$.
• Initial Check: As $x \to 7$, the numerator approaches $\sqrt{7+2} - 3 = \sqrt{9} - 3 = 3 - 3 = 0$.
• The denominator approaches $7 - 7 = 0$.
• Indeterminate Form: This is a $0/0$ indeterminate form, meaning limit laws for quotients cannot be applied directly due to division by zero.
  • Indeterminate: The overall behavior (approaching zero, infinity, or a finite number) cannot be determined immediately from the $0/0$ form.
• Strategy: Simplify the expression to eliminate the problematic factors (hidden $(x-7)$ in the numerator).
• Technique: Multiply by the conjugate of the numerator ($\sqrt{x+2} + 3$).
  • $ext{lim}_{x \to 7} \frac{\sqrt{x+2} - 3}{x-7} \cdot \frac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3}$
• Expansion: Using $(a-b)(a+b) = a^2 - b^2$ for the numerator:
  • Numerator: $(\sqrt{x+2})^2 - 3^2 = (x+2) - 9 = x-7$.
  • Original expression becomes: $ext{lim}_{x \to 7} \frac{x-7}{(x-7)(\sqrt{x+2} + 3)}$
• Cancellation: Since $x \to 7$ means $x \neq 7$, we can cancel the $(x-7)$ terms.
  • $ext{lim}_{x \to 7} \frac{1}{\sqrt{x+2} + 3}$
• Apply Limit Laws: Now, apply limit laws to the simplified expression.
  • Numerator: $ext{lim}_{x \to 7} 1 = 1$.
  • Denominator: $ext{lim}_{x \to 7} (\sqrt{x+2} + 3) = \sqrt{7+2} + 3 = \sqrt{9} + 3 = 3 + 3 = 6$.
• Result: $\frac{1}{6}$.

Example 3: Indeterminate Form and Factoring

• Problem: Evaluate $ext{lim}_{x \to 1} \frac{x^2 - 1}{x-1}$.
• Initial Check: Both numerator ($1^2 - 1 = 0$) and denominator ($1-1 = 0$) approach zero ($0/0$ indeterminate form).
• Strategy: Factor the numerator using the difference of squares formula ($a^2 - b^2 = (a-b)(a+b)$).
  • $x^2 - 1 = (x-1)(x+1)$.
• Expression becomes: $ext{lim}_{x \to 1} \frac{(x-1)(x+1)}{x-1}$
• Cancellation: Since $x \neq 1$, cancel $(x-1)$ terms.
  • $ext{lim}_{x \to 1} (x+1)$
• Apply Limit Laws: $1 + 1 = 2$.

Example 4: Indeterminate Form and Expansion/Simplification

• Problem: Evaluate $ext{lim}_{x \to 0} \frac{(3+x)^2 - 3^2}{x}$.
• Initial Check: Both numerator ($(3+0)^2 - 3^2 = 9 - 9 = 0$) and denominator ($0$) approach zero ($0/0$ indeterminate form).
• Strategy: Expand the quadratic term in the numerator.
  • $(3+x)^2 = 3^2 + 2(3)(x) + x^2 = 9 + 6x + x^2$.
• Expression becomes: $ext{lim}<em>{x \to 0} \frac{(9 + 6x + x^2) - 9}{x} = ext{lim}</em>{x \to 0} \frac{6x + x^2}{x}$
• Simplification: Factor out $x$ from the numerator or split the fraction.
  • $ext{lim}<em>{x \to 0} (\frac{6x}{x} + \frac{x^2}{x}) = ext{lim}</em>{x \to 0} (6 + x)$
• Apply Limit Laws: $6 + 0 = 6$.

Squeeze Theorem for Functions

• Context: Used when limit laws do not apply, and algebraic simplification is not possible (e.g., limits involving oscillating functions like $ext{sin}(1/x)$).
• Scenario: Consider $ext{lim}_{x \to 0} x \text{sin}(1/x)$.
  • $ext{lim}_{x \to 0} x = 0$.
  • $ext{lim}_{x \to 0} \text{sin}(1/x)$ does not exist (DNE) because as $x \to 0$, $1/x \to \pm \infty$, causing $\text{sin}(1/x)$ to oscillate rapidly between $-1$ and $1$. Therefore, the product limit law cannot be directly applied.
• Statement: If, for $x$ in an interval around $a$ (but not necessarily at $a$),
  $g(x) \le f(x) \le h(x)$
  and if $ext{lim}<em>{x \to a} g(x) = L$ and $ext{lim}</em>{x \to a} h(x) = L$,
  then $ext{lim}_{x \to a} f(x) = L$.
• Intuition: If $f(x)$ is