Calculus Study Notes: Functions, Limits, and Theorems

Functions and Their Domains

Finding the Domain of a Function

The domain of a function is the set of all possible input values (usually x-values) for which the function is defined. For functions involving square roots and rational expressions, the domain is restricted by the requirement that the expression under the square root must be non-negative and the denominator must not be zero.

• Example: For $f(x) = \sqrt{x^2 - 25x}$, set $x^2 - 25x \geq 0$ to find the domain.

• Steps:

  1. Solve $x^2 - 25x \geq 0$.

  2. Factor: $x(x - 25) \geq 0$.

  3. Test intervals: $x \leq 0$ or $x \geq 25$.

  4. Domain: $(-\infty, 0] \cup [25, \infty)$

• Example: For $f(x) = \frac{1}{x^2 - 5x}$, set $x^2 - 5x \neq 0$.

• Steps:

  1. Solve $x^2 - 5x = 0 \implies x(x-5) = 0 \implies x = 0$ or $x = 5$.

  2. Domain: $(-\infty, 0) \cup (0, 5) \cup (5, \infty)$

Composite Functions

Finding and Simplifying Composite Functions

A composite function is formed when one function is substituted into another. The notation $f(g(x))$ means to substitute $g(x)$ into every occurrence of $x$ in $f(x)$.

• Given: $f(x) = 2x - 1$, $g(x) = 3x^2 + 5$

• Find: $f(g(x))$

  1. Substitute $g(x)$ into $f(x)$: $f(g(x)) = 2(3x^2 + 5) - 1$

  2. Simplify: $= 6x^2 + 10 - 1 = 6x^2 + 9$

• Find: $g(f(x))$

  1. Substitute $f(x)$ into $g(x)$: $g(f(x)) = 3(2x - 1)^2 + 5$

  2. Expand: $(2x - 1)^2 = 4x^2 - 4x + 1$

  3. Simplify: $= 3(4x^2 - 4x + 1) + 5 = 12x^2 - 12x + 3 + 5 = 12x^2 - 12x + 8$

Limits and Analytical Evaluation

Finding Limits Analytically

The limit of a function as $x$ approaches a value describes the behavior of the function near that point. Analytical methods include direct substitution, factoring, and rationalizing.

• Example 1: $\lim_{x \to 7} \frac{x-7}{7x}$

  • Substitute $x = 7$: $\frac{7-7}{7 \times 7} = \frac{0}{49} = 0$

• Example 2: $\lim_{x \to 1} \frac{4x^2 - 1}{x^2 - 1}$

  • Factor numerator and denominator: $4x^2 - 1 = (2x+1)(2x-1)$, $x^2 - 1 = (x+1)(x-1)$

  • Substitute $x = 1$: $\frac{(2 \times 1 + 1)(2 \times 1 - 1)}{(1+1)(1-1)} = \frac{3 \times 1}{2 \times 0} = \frac{3}{0}$ (undefined)

  • Check for removable discontinuity or use L'Hospital's Rule if necessary.

• Example 3: $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

  • Factor numerator: $x^2 - 9 = (x-3)(x+3)$

  • Cancel $(x-3)$: $\frac{(x-3)(x+3)}{x-3} = x+3$

  • Substitute $x = 3$: $3 + 3 = 6$

• Example 4: $\lim_{x \to 5} \frac{x^2 - 25}{x - 5}$

  • Factor numerator: $x^2 - 25 = (x-5)(x+5)$

  • Cancel $(x-5)$: $\frac{(x-5)(x+5)}{x-5} = x+5$

  • Substitute $x = 5$: $5 + 5 = 10$

Limits Involving Trigonometric Functions

• Example: $\lim_{x \to 0} \frac{\sin(2x) \sec(x)}{x}$

  • Recall $\sec(x) = \frac{1}{\cos(x)}$

  • Rewrite: $\frac{\sin(2x)}{x \cos(x)}$

  • As $x \to 0$, $\sin(2x) \approx 2x$, $\cos(x) \approx 1$

  • Limit: $\frac{2x}{x \cdot 1} = 2$

• Example: $\lim_{x \to 0} \frac{\sin(3x)}{x \cos(x)}$

  • As $x \to 0$, $\sin(3x) \approx 3x$, $\cos(x) \approx 1$

  • Limit: $\frac{3x}{x \cdot 1} = 3$

• Example: $\lim_{x \to 0} \frac{\cos(x) - 1}{x}$

  • As $x \to 0$, $\cos(x) \to 1$, so numerator $\to 0$

  • Use L'Hospital's Rule: Derivative of numerator $= -\sin(x)$, denominator $= 1$

  • Limit: $\lim_{x \to 0} \frac{-\sin(x)}{1} = 0$

Intermediate Value Theorem (IVT)

Statement and Application

The Intermediate Value Theorem states that if a function $f$ is continuous on a closed interval $[a, b]$, and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c$ in $[a, b]$ such that $f(c) = N$.

• Steps to Apply IVT:

  1. Check that $f(x)$ is continuous on $[a, b]$.

  2. Calculate $f(a)$ and $f(b)$.

  3. If $f(a)$ and $f(b)$ have opposite signs, then there is at least one root in $[a, b]$.

• Example: Show that $f(x) = x^3 + x^2 - 9x - 9$ has a solution in $[-5, 5]$.

  • Check continuity: $f(x)$ is a polynomial, so it is continuous everywhere.

  • Calculate $f(-5)$ and $f(5)$:

  • $f(-5) = (-5)^3 + (-5)^2 - 9(-5) - 9 = -125 + 25 + 45 - 9 = -64$

  • $f(5) = 125 + 25 - 45 - 9 = 96$

  • Since $f(-5) < 0$ and $f(5) > 0$, by IVT, there is at least one root in $[-5, 5]$.

Formal Definition of the Limit (Epsilon-Delta)

Using Epsilon-Delta to Prove Limits

The formal definition of the limit uses two quantities, $\epsilon$ (epsilon) and $\delta$ (delta), to rigorously define what it means for $\lim_{x \to a} f(x) = L$.

• Definition: For every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$.

• Example: Prove $\lim_{x \to 4} 7x - 8 = 20$.

  • Let $|7x - 8 - 20| < \epsilon$

  • Simplify: $|7x - 28| = 7|x - 4| < \epsilon$

  • So, $|x - 4| < \frac{\epsilon}{7}$

  • Choose $\delta = \frac{\epsilon}{7}$

Summary Table: Key Concepts

Concept	Definition	Example
Domain of Function	Set of all input values for which the function is defined	$f(x) = \sqrt{x-2}$, domain: $x \geq 2$
Composite Function	Function formed by substituting one function into another	$f(g(x))$
Limit	Value that a function approaches as the input approaches a point	$\lim_{x \to 3} (x+2) = 5$
Intermediate Value Theorem	If $f$ is continuous on $[a, b]$, then $f$ takes every value between $f(a)$ and $f(b)$	Root exists if $f(a)$ and $f(b)$ have opposite signs
Epsilon-Delta Definition	Formal definition of the limit using $\epsilon$ and $\delta$	$|f(x) - L| < \epsilon$ when $|x - a| < \delta$

Additional info: Some steps and explanations have been expanded for clarity and completeness, including the use of L'Hospital's Rule and the formal definition of the limit.