Unit 13 Precalculus Practice: Continuity, Squeeze Theorem, and Derivatives

M.O. #13.1 - Continuity and Discontinuity

• The Three Conditions for Continuity at a Point ($x = a$):     To prove that a function is continuous at a specific point $x = a$, three conditions must hold. If any of these conditions are not met, the function is immediately concluded to be discontinuous at that point:     * Condition 1: $f(a)$ exists.     * Condition 2: $\lim_{x \to a} f(x)$ exists.     * Condition 3: $\lim_{x \to a} f(x) = f(a)$.

• Removable Discontinuity:     To show that a function has a removable discontinuity at $x = a$, it must be shown that the limit exists at that point: $\lim_{x \to a} f(x)$ exists.

• Example 1: Discontinuity at $x = 4$ (Failure of Condition #2):     * Given values: $\lim_{x \to 4^-} f(x) = 5$ and $\lim_{x \to 4^+} f(x) = 8$.     * Explanation: Since the left-hand limit and right-hand limit are not equal, $\lim_{x \to 4} f(x) = \text{DNE}$ (Does Not Exist). This fails Condition #2.

• Example 2: Discontinuity at $x = 3$ (Failure of Condition #3):     * Given values: $f(3) = 4$ and $\lim_{x \to 3} f(x) = 6$.     * Explanation: Even though both the value and the limit exist, they are not equal ($f(3) \neq \lim_{x \to 3} f(x)$). This fails Condition #3.

• Example 3: Proving Continuity at $x = 3$:     * Function: $f(x) = \begin{cases} \sqrt{x+1} & \text{if } x \leq 3 \\ 5-x & \text{if } x > 3 \end{cases}$     * Verification step 1 ($f(a)$): $f(3) = \sqrt{3+1} = \sqrt{4} = 2$. (Exists)     * Verification step 2 ($\lim f(x)$):         * $\lim_{x \to 3^-} f(x) = \sqrt{3+1} = 2$         * $\lim_{x \to 3^+} f(x) = 5-3 = 2$         * Therefore, $\lim_{x \to 3} f(x) = 2$. (Exists)     * Verification step 3 ($\lim f(x) = f(a)$): $2 = 2$. (Matches)     * Conclusion: The function is continuous at $x = 3$.

• Example 4: Identifying Discontinuities for $f(x) = \frac{x+3}{x^2-9}$:     * Discontinuities occur where the denominator is zero: $x^2 - 9 = 0 \rightarrow (x+3)(x-3) = 0$.     * Values: Discontinuous at $x = -3$ and $x = 3$.     * Removable Discontinuity: The discontinuity at $x = -3$ is removable because the factor $(x+3)$ can be cancelled from the numerator and denominator.     * Showing the limit for removable discontinuity:         $\lim_{x \to -3} \frac{x+3}{(x+3)(x-3)} = \lim_{x \to -3} \frac{1}{x-3} = \frac{1}{-3-3} = -\frac{1}{6}$

• Example 5: Solving for a Constant ($k$) to Ensure Continuity:     * Function: $f(x) = \begin{cases} \sqrt{x}+6 & \text{if } x < 9 \\ k-2x & \text{if } x \geq 9 \end{cases}$     * Set limits equal at $x = 9$:         $\sqrt{9} + 6 = k - 2(9)$$3 + 6 = k - 18$$9 = k - 18$$k = 27$

M.O. #13.2 - Squeeze Theorem

• Definition:     If $g(x) \leq f(x) \leq h(x)$ and $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$.

• Example 6: Using Squeeze Theorem:     * Given: $-2x^3 \leq f(x) \leq x+3$. Find $\lim_{x \to -1} f(x)$.     * Evaluate boundary limits:         $\lim_{x \to -1} (-2x^3) = -2(-1)^3 = -2(-1) = 2$$\lim_{x \to -1} (x+3) = -1 + 3 = 2$     * Conclusion: Since both outside limits equal $2$, $\lim_{x \to -1} f(x) = 2$.

• Example 7: Showing $\lim_{x \to 0} (x^3 \cos(\frac{1}{x})) = 0$:     * Start with the known range of the cosine function: $-1 \leq \cos(\frac{1}{x}) \leq 1$.     * Multiply throughout by $x^3$: $-x^3 \leq x^3 \cos(\frac{1}{x}) \leq x^3$.     * Determine limits of the bounding functions as $x \to 0$:         $\lim_{x \to 0} (-x^3) = -(0)^3 = 0$$\lim_{x \to 0} (x^3) = (0)^3 = 0$     * Conclusion: By Squeeze Theorem, $\lim_{x \to 0} (x^3 \cos(\frac{1}{x})) = 0$.

M.O. #13.3 - Limit to Find Slope of a Tangent Line / Instantaneous Rate of Change

• Points of Interest: $(x, f(x))$ and $(x_1, f(x_1))$.

• Slope Formula for Tangent Line:$m = \lim_{x_1 \to x} \frac{f(x_1) - f(x)}{x_1 - x}$

• Example 8: Analyzing $f(x) = x^2 + 2$:     * Part a: Secant Line ($x = 3$ to $x = 5$):         * Points: $(3, 11)$ and $(5, 27)$.         * Slope Calculation: $m_{\text{sec}} = \frac{27 - 11}{5 - 3} = \frac{16}{2} = 8$.         * Type: This represents the average rate of change.         * Equation: $y - 11 = 8(x - 3)$.     * Part b: Tangent Line (at $x = 4$):         * Point: $(4, 18)$.         * Slope formula: $m_{\text{tan}} = \lim_{x \to 4} \frac{x^2 + 2 - 18}{x - 4} = \lim_{x \to 4} \frac{x^2 - 16}{x - 4}$.         * Factor: $\lim_{x \to 4} \frac{(x+4)(x-4)}{x-4} = \lim_{x \to 4} (x+4) = 4 + 4 = 8$.         * Type: This represents the instantaneous rate of change.         * Equation: $y - 18 = 8(x - 4)$.     * Part c: General Expression for Slope at $(x_0, f(x_0))$:$m_{\text{tan}} = \lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)}{x_1 - x_0}$$m_{\text{tan}} = \lim_{x_1 \to x_0} \frac{x_1^2 + 2 - (x_0^2 + 2)}{x_1 - x_0}$$m_{\text{tan}} = \lim_{x_1 \to x_0} \frac{x_1^2 - x_0^2}{x_1 - x_0} = \lim_{x_1 \to x_0} (x_1 + x_0) = x_0 + x_0 = 2x$

M.O. #13.4 - Definition of the Derivative (First Principles)

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

• Example 9: Given $f(x) = 2x^2 - 3$:     * Part a: Find $f'(x)$:$f'(x) = \lim_{h \to 0} \frac{2(x+h)^2 - 3 - (2x^2 - 3)}{h}$$f'(x) = \lim_{h \to 0} \frac{2(x^2 + 2xh + h^2) - 3 - 2x^2 + 3}{h}$$f'(x) = \lim_{h \to 0} \frac{2x^2 + 4xh + 2h^2 - 2x^2}{h}$$f'(x) = \lim_{h \to 0} \frac{4xh + 2h^2}{h} = \lim_{h \to 0} (4x + 2h)$$f'(x) = 4x + 2(0) = 4x$     * Part b: Tangent Line at $x = -2$:         * Slope: $m_{\text{tan}} = 4(-2) = -8$.         * Point: $f(-2) = 2(-2)^2 - 3 = 8 - 3 = 5$. Point is $(-2, 5)$.         * Equation: $y - 5 = -8(x + 2)$.

M.O. #13.5 - Interpreting Derivative Limits

• Example 10: Recognizing Limits:     * a.) $\lim_{x \to 2} \frac{f(x) - f(2)}{x-2}$: Derivative of $f(x)$ at $x = 2$.     * b.) $\lim_{x \to 0} \frac{\ln(x) - 1}{x}$: Derivative of $f(x) = \ln(x)$ at $x = e$ (Note: Logic dictates this refers to the definition at a specific point where $f(x)=1$).     * c.) $\lim_{h \to 0} \frac{(x+h)^4 - x^4}{h}$: Derivative of $f(x) = x^4$.     * d.) $\lim_{h \to 0} \frac{\sqrt{25+h} - 5}{h}$: Derivative of $f(x) = \sqrt{x}$ at $x = 25$.

• Example 11: General Tangent Equation:     * Given: $f(-2) = 3$ and $f'(-2) = 4$.     * Slope = $4$, Point = $(-2, 3)$.     * Equation: $y - 3 = 4(x + 2)$.

• Example 12: Horizontal Tangents:     * If $f'(6) = 0$ and $f(6) = -1$, the line tangent to $f(x)$ at $x = 6$ will be horizontal and have a y-intercept of $(0, -1)$.

• Example 13: Tangent line to $f(x) = \frac{5}{x} + 1$ at $x = 2$:     * $f(2) = \frac{5}{2} + 1 = 3.5$. Point is $(2, 3.5)$.     * $f'(2) = \lim_{h \to 0} \frac{\frac{5}{2+h} + 1 - (\frac{5}{2} + 1)}{h} = \dots = -\frac{5}{4}$.     * Equation: $y - 3.5 = -1.25(x - 2)$.

• Example 14: Tangent line to $f(x) = 5 - \sqrt{x}$ at $x = 9$:     * $f(9) = 5 - \sqrt{9} = 2$. Point is $(9, 2)$.     * Slope calculation using limit results in $m = -\frac{1}{6}$.     * Equation: $y - 2 = -\frac{1}{6}(x - 9)$.