Calculus: Limits, Continuity, and the Limit Definition of Derivatives

Derivative of a Constant Function: For the function $f(x) = -3$ , the derivative is $f'(x) = 0$ . - Mathematical Justification: The graph of a constant function is a horizontal line. Since the derivative represents the slope of the tangent line at any point, and the slope of any horizontal line is always zero, the derivative must be $0$ .

Function Analysis: Given the function $f(x) = \frac{1}{x-3}$ .
Calculation of the Derivative using the Limit Definition: - The derivative is found using the difference quotient: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ . - Substituting the function: $f'(x) = \frac{\frac{1}{(x+h)-3} - \frac{1}{x-3}}{h}$ . - Finding a common denominator for the numerator: $\frac{(x-3) - (x+h-3)}{(x+h-3)(x-3)(h)}$ . - Simplifying the numerator: $x - 3 - x - h + 3 = -h$ . - The expression becomes: $\frac{-h}{h(x+h-3)(x-3)}$ . - Canceling the $h$ terms results in: $\frac{-1}{(x+h-3)(x-3)}$ . - Taking the limit as $h \to 0$ : $f'(x) = \frac{-1}{(x-3)^2}$ .
Equation of the Tangent Line at $x = 5$ : - Step 1: Find the y-coordinate. Evaluate $f(5)$ : $f(5) = \frac{1}{5-3} = \frac{1}{2}$ . The point of tangency is $(5, \frac{1}{2})$ . - Step 2: Find the slope. Evaluate $f'(5)$ : $f'(5) = \frac{-1}{(5-3)^2} = -\frac{1}{4}$ . - Step 3: Point-Slope Form. Use $y - y_1 = m(x - x_1)$ : $y - \frac{1}{2} = -\frac{1}{4}(x - 5)$ . - Step 4: Solve for y. - $y - \frac{1}{2} = -\frac{1}{4}x + \frac{5}{4}$ . - $y = -\frac{1}{4}x + \frac{5}{4} + \frac{2}{4}$ . - Final equation: $y = -\frac{1}{4}x + \frac{7}{4}$ .\n

Function Analysis: Given $f(x) = x^3 - 12x$ .
Finding the Derivative Algebraicly: - Difference Quotient: $\frac{(x+h)^3 - 12(x+h) - (x^3 - 12x)}{h}$ . - Expanding terms: $\frac{x^3 + 3x^2h + 3xh^2 + h^3 - 12x - 12h - x^3 + 12x}{h}$ . - Simplifying: $\frac{3x^2h + 3xh^2 + h^3 - 12h}{h}$ . - Factoring out $h$ : $\frac{h(3x^2 + 3xh + h^2 - 12)}{h}$ . - Taking the limit as $h \to 0$ : $f'(x) = 3x^2 - 12$ .
Determining Points of Horizontal Tangency: - A tangent line is horizontal when the derivative (slope) is zero. - Set $f'(x) = 0$ : $3x^2 - 12 = 0 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2$ . - Evaluating Coordinates: - For $x = 2$ : $f(2) = (2)^3 - 12(2) = 8 - 24 = -16$ . Point: $(2, -16)$ . - For $x = -2$ : $f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16$ . Point: $(-2, 16)$ .

Based on the analysis of the provided graph of $f(x)$ , the following limits are determined:

Individual Limit Points: - $\lim_{x \to 7} f(x) = 1$ . - $\lim_{x \to 5^+} f(x) = 8$ . - $\lim_{x \to 5^-} f(x) = 3$ . - $\lim_{x \to 5} f(x) = \text{DNE}$ (Since left and right hand limits are unequal: $3 \neq 8$ ). - $\lim_{x \to 2^-} f(x) = \text{DNE}$ . - $\lim_{x \to 11} f(x) = 4$ . - $\lim_{x \to \infty} f(x) = 4$ .

Problem #9 Data Analysis: - Tables show values approaching 5 from both sides: $2.9, 2.99, 2.999$ and $3.001, 3.01, 3.1$ . - At exactly $x=5$ , the calculator shows "ERROR". - Limit Value: $\lim_{x \to 5} f(x) = 3$ .
Problem #10 Data Analysis: - Tables show the same approach: $2.9, 2.9, …$ and $…, 3.01, 3.1$ . - At exactly $x=5$ , the output is defined as $3$ . - Limit Value: $\lim_{x \to 5} f(x) = 3$ .
Comparison of Findings: - Problem #9 indicates a hole at the coordinate $(5, 3)$ because the limit exists but the function is undefined at that specific point. - Problem #10 indicates a point exists at $(5, 3)$ because the function is defined and continuous at that value.
Additional Table Examples: - Problem #11: Values approach $8$ , but the function value at $x=5$ is defined as $12$ . The limit is $8$ . - Problem #12: Left-hand side approaches $3$ , right-hand side approaches $9$ . The limit does not exist (DNE).

Compound Fraction Evaluation: - $\lim_{x \to 0} \frac{\frac{1}{3} - \frac{1}{x+3}}{x}$ . - Simplify numerator: $\frac{(x+3) - 3}{3(x+3)} = \frac{x}{3(x+3)}$ . - Divide by $x$ : $\frac{x}{x \cdot 3(x+3)} = \frac{1}{3(x+3)}$ . - Substitute $x=0$ : $\frac{1}{3(0+3)} = \frac{1}{9}$ .
Factoring Rational Functions: - $\lim_{x \to 4} \frac{4-x}{x^2-16} = \lim_{x \to 4} \frac{-(x-4)}{(x-4)(x+4)}$ . - Cancel $(x-4)$ : $\lim_{x \to 4} \frac{-1}{x+4} = -\frac{1}{8}$ .
Conjugate Method for Radicals: - $\lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x}$ . - Multiply by conjugate $\frac{\sqrt{x+3} + \sqrt{3}}{\sqrt{x+3} + \sqrt{3}}$ . - Numerator becomes $(x+3) - 3 = x$ . - Expression: $\frac{x}{x(\sqrt{x+3} + \sqrt{3})} = \frac{1}{\sqrt{x+3} + \sqrt{3}}$ . - Substitute $x=0$ : $\frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}$ .
Piecewise Function Limits: - Given $f(x) = \begin{cases} 3x-1, & x \leq 4 \ 2x+1, & x > 4 \end{cases}$ . - Left-hand limit: $3(4)-1 = 11$ . - Right-hand limit: $2(4)+1 = 9$ . - Conclusion: $\lim_{x \to 4} f(x) = \text{DNE}$ because $11 \neq 9$ .
Limits at Infinity: - $\lim_{x \to \infty} \frac{5x-2}{3x+2}$ . Since the degree of the numerator and denominator are the same, the limit is the ratio of leading coefficients: $\frac{5}{3}$ . - $\lim_{x \to \infty} \frac{4x^3-2x^2+6x}{-3x^2-5x+1}$ . Since the degree of the numerator is higher than the denominator, the limit is $\text{DNE}$ (it approaches infinity).

Slope of a Parabola at a Point: - Function: $f(x) = 3x^2 - 8$ at the point $(4, 40)$ . - Use difference quotient: $\frac{3(x+h)^2 - 8 - (3x^2 - 8)}{h}$ . - Expand: $\frac{3(x^2 + 2xh + h^2) - 8 - 3x^2 + 8}{h} = \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h}$ . - Simplify: $\frac{6xh + 3h^2}{h} = 6x + 3h$ . - Take limit $h \to 0$ : $f'(x) = 6x$ . - Evaluate at $x=4$ : $6(4) = 24$ . The slope at $(4, 40)$ is $24$ .
Derivative of a Square Root Function: - Function: $f(x) = \sqrt{x-1}$ . - Difference Quotient: $\frac{\sqrt{x+h-1} - \sqrt{x-1}}{h}$ . - Multiply by conjugate $\frac{\sqrt{x+h-1} + \sqrt{x-1}}{\sqrt{x+h-1} + \sqrt{x-1}}$ . - Numerator: $(x+h-1) - (x-1) = h$ . - Denominator: $h(\sqrt{x+h-1} + \sqrt{x-1})$ . - Cancel $h$ : $\frac{1}{\sqrt{x+h-1} + \sqrt{x-1}}$ . - Take limit $h \to 0$ : $f'(x) = \frac{1}{2\sqrt{x-1}}$ .