Business Calculus: Limits, Continuity, and Differentiation (MAT 114 Test 1 Review)

Limits

Understanding Limits

Limits are a foundational concept in calculus, describing the behavior of a function as the input approaches a particular value. They are essential for defining continuity and derivatives.

• Definition: The limit of a function f(x) as x approaches a is the value that f(x) gets closer to as x gets closer to a.

• Notation: $\lim_{x \to a} f(x)$

• Example: For the function shown in the graph, $\lim_{x \to -2} f(x) = 1$ and $\lim_{x \to 2} f(x) = 1$.

• One-sided limits: $\lim_{x \to 0^-} f(x) = -2$, $\lim_{x \to 0^+} f(x)$ does not exist (DNE).

• Direct substitution: If plugging in the value gives a real number, the limit equals the function value.

Evaluating Limits Algebraically

When direct substitution leads to an indeterminate form (such as $\frac{0}{0}$), algebraic manipulation is required.

• Example: $\lim_{x \to 4} \frac{x^2 - 4x}{x^2 - 6x + 8}$

Step-by-step solution:

• Substitute $x = 4$: $\frac{4^2 - 4(4)}{4^2 - 6(4) + 8} = \frac{0}{0}$ (indeterminate).

• Factor numerator: $x^2 - 4x = x(x - 4)$

• Factor denominator: $x^2 - 6x + 8 = (x - 2)(x - 4)$

• Simplify: $\frac{x(x - 4)}{(x - 2)(x - 4)} = \frac{x}{x - 2}$

• Take the limit: $\lim_{x \to 4} \frac{x}{x - 2} = \frac{4}{2} = 2$

Continuity

Continuity at a Point

A function is continuous at a point if its limit at that point equals its value at that point.

• Definition: f is continuous at x = a if and only if $\lim_{x \to a} f(x) = f(a)$.

• Three conditions for continuity at a:

  1. f(a) is defined.

  2. $\lim_{x \to a} f(x)$ exists.

  3. $\lim_{x \to a} f(x) = f(a)$.

• If any condition fails, f is discontinuous at a.

• If f is continuous at every point in an interval, it is continuous on that interval.

Example: Piecewise Function Continuity

Consider the function:

• $h(x) = \begin{cases} 3x - 1 & \text{if } x < 2 \\ 7 - x & \text{if } x > 2 \\ 6 & \text{if } x = 2 \end{cases}$

To check continuity at $x = 2$:

• $\lim_{x \to 2^-} h(x) = 3(2) - 1 = 5$

• $\lim_{x \to 2^+} h(x) = 7 - 2 = 5$

• $\lim_{x \to 2} h(x) = 5$

• $h(2) = 6$

• Since $\lim_{x \to 2} h(x) \neq h(2)$, $h(x)$ is not continuous at $x = 2$.

Derivatives

The Derivative Function

The derivative of a function at a point measures the instantaneous rate of change, or the slope of the tangent line at that point.

• Definition: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

• If $f'(x)$ exists at $x$, $f$ is differentiable at $x$.

• The process of finding a derivative is called differentiation.

Example: Derivative by Definition

• Given $f(x) = \frac{1}{x + 2}$, find $f'(x)$.

Solution:

• $f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h + 2} - \frac{1}{x + 2}}{h}$

• Simplify numerator: $\frac{1}{x + h + 2} - \frac{1}{x + 2} = \frac{-(h)}{(x + h + 2)(x + 2)}$

• So, $f'(x) = \lim_{h \to 0} \frac{-h}{h(x + h + 2)(x + 2)} = \lim_{h \to 0} \frac{-1}{(x + h + 2)(x + 2)} = -\frac{1}{(x + 2)^2}$

Differentiation Rules

Basic Rules Summary

Several rules simplify the process of differentiation for common functions.

• Constant Rule: $\frac{d}{dx} c = 0$

• Power Rule: $\frac{d}{dx} x^n = n x^{n-1}$

• Constant Multiple Rule: $\frac{d}{dx} [c f(x)] = c f'(x)$

• Sum Rule: $\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)$

• Difference Rule: $\frac{d}{dx} [f(x) - g(x)] = f'(x) - g'(x)$

Example: Polynomial and Radical Functions

• Given $g(w) = 8w^3 - 18w^2 + 7w - 5$, differentiate term-by-term:

• $g'(w) = 24w^2 - 36w + 7$

• For radicals, rewrite as powers: $\sqrt{y} = y^{1/2}$

• Example: $f(y) = 6y^{1/2} - 8y^{3/2}$

• $f'(y) = 6 \cdot \frac{1}{2} y^{-1/2} - 8 \cdot \frac{3}{2} y^{1/2} = 3y^{-1/2} - 12y^{1/2}$

• For $h(x) = x^{3/2} + 2x - x^{1/2} - 2$, $h'(x) = \frac{3}{2} x^{1/2} + 2 - \frac{1}{2} x^{-1/2}$

Product and Quotient Rules

Product Rule

The product rule is used to differentiate products of two functions.

• Formula: $(uv)' = u'v + uv'$

• Example: For $y = (x^2 + 1)(\sqrt{x} + 1)$, let $u(x) = x^2 + 1$, $v(x) = x^{1/2} + 1$

• Apply product rule: $\frac{dy}{dx} = u'v + uv'$

• $u'(x) = 2x$, $v'(x) = \frac{1}{2} x^{-1/2}$

• $\frac{dy}{dx} = 2x(x^{1/2} + 1) + (x^2 + 1) \cdot \frac{1}{2} x^{-1/2}$

• Simplify: $2x x^{1/2} + 2x + \frac{1}{2} x^2 x^{-1/2} + \frac{1}{2} x^{-1/2}$

• Final answer: $\frac{5}{2} x^{3/2} + 2x + \frac{1}{2} x^{-1/2}$

Quotient Rule

The quotient rule is used to differentiate quotients of two functions.

• Formula: $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$

• Example: For $f(x) = \frac{x + 1}{x + 10}$, $u(x) = x + 1$, $v(x) = x + 10$

• Apply quotient rule: $f'(x) = \frac{(x + 1)'(x + 10) - (x + 1)(x + 10)'}{(x + 10)^2}$

• Compute derivatives: $(x + 1)' = 1$, $(x + 10)' = 1$

• Substitute: $f'(x) = \frac{1 \cdot (x + 10) - (x + 1) \cdot 1}{(x + 10)^2} = \frac{x + 10 - x - 1}{(x + 10)^2} = \frac{9}{(x + 10)^2}$

Chain Rule

Chain Rule for Composite Functions

The chain rule is used to differentiate composite functions, where one function is nested inside another.

• Formula: If $y = f(u)$ and $u = g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

• Alternatively, $[f(g(x))]' = f'(g(x)) \cdot g'(x)$

• Example: For $y = (x^2 + x + 4)^{1/2}$, let $u = x^2 + x + 4$ (inner function), $y = u^{1/2}$ (outer function)

• $\frac{dy}{du} = \frac{1}{2} u^{-1/2}$, $\frac{du}{dx} = 2x + 1$

• By chain rule: $\frac{dy}{dx} = \frac{1}{2} (x^2 + x + 4)^{-1/2} \cdot (2x + 1) = \frac{2x + 1}{2(x^2 + x + 4)^{1/2}}$

Summary Table: Differentiation Rules

The following table summarizes the main differentiation rules covered:

Rule	Formula	Example
Constant Rule	$\frac{d}{dx} c = 0$	$\frac{d}{dx} 5 = 0$
Power Rule	$\frac{d}{dx} x^n = n x^{n-1}$	$\frac{d}{dx} x^3 = 3x^2$
Sum Rule	$\frac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x)$	$\frac{d}{dx} (x^2 + x) = 2x + 1$
Product Rule	$(uv)' = u'v + uv'$	$\frac{d}{dx} (x^2 \cdot \sin x)$
Quotient Rule	$\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$	$\frac{d}{dx} \left( \frac{x}{x+1} \right)$
Chain Rule	$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$	$\frac{d}{dx} (\sin(x^2))$

Additional info: These notes are based on a review for MAT 114 (Business Calculus), covering limits, continuity, and differentiation rules, with examples and step-by-step solutions. The content is suitable for college students preparing for a Business Calculus exam.