Calculus Notes: Algebraic Simplification, Trigonometric Limits, and One-Sided Discontinuities

Direct Substitution and Algebraic Simplification of Limits

• Naive Concept of Continuity:

  • A continuous function can be understood intuitively as a curve that can be drawn in a single continuous motion without lifting the writing instrument off the page.
  • If a function $g(x)$ has a removable discontinuity at $x = 3$ where its limit is $6$, defining or redefining $g(3) = 6$ eliminates the hole and makes the function continuous at that point.

• General Strategy for Calculating Limits:

  • Always begin evaluating any limit by attempting direct substitution of the target value for the variable.
  • If direct substitution yields a defined, real number, that result is the limit.
  • If direct substitution results in an indeterminate form (such as $\frac{0}{0}$), algebraic manipulation must be performed to simplify the expression before evaluating the limit.

• Instantaneous Velocity and Difference Quotients:

  • Instantaneous velocity at time $t$ is defined as the limit of average velocity over a time interval $[t, t+h]$ as $h$ approaches $0$.
  • For an object (Sparky the dragon bicycling away from Eel) with position function $f(t) = 2t^2 + 10$, calculating the instantaneous velocity at time $t = 1\text{\,s}$ involves evaluating:         $\lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$

• Step-by-Step Algebraic Simplification of the Difference Quotient:

  • Step 1: Set up the quotient using the position function:$\frac{(2(1+h)^2 + 10) - (2(1)^2 + 10)}{h}$
  • Step 2: Distribute the negative sign in the second term:$-(2(1)^2 + 10) = -2(1) - 10 = -2 - 10$
    • The constant term $+10$ in the first term cancels with $-10$ from the second term.
    • This leaves the simplified numerator:             $2(1+h)^2 - 2$
  • Step 3: Expand the binomial term:$(1+h)^2 = 1 + 2h + h^2$
  • Step 4: Distribute the coefficient $2$ across the binomial:$2(1 + 2h + h^2) = 2 + 4h + 2h^2$
  • Step 5: Subtract the remaining constant $2$:$(2 + 4h + 2h^2) - 2 = 4h + 2h^2$
  • Step 6: Reconstruct the simplified difference quotient:$\frac{4h + 2h^2}{h}$
  • Step 7: Factor out $h$ from the numerator to remove division by zero:$\frac{h(4 + 2h)}{h}$
  • Step 8: Cancel the factor $h$ in both numerator and denominator:$4 + 2h$
  • Step 9: Direct substitution of $h = 0$ into the remaining linear expression:$\lim_{h \to 0} (4 + 2h) = 4 + 2(0) = 4$
  • Final Result: The instantaneous velocity at $t = 1\text{\,s}$ is $4\,\text{m/s}$.
  • Warning on common errors: Performing these precise algebraic steps correctly is critical; early structural errors prevent the cancellation of $h$ in the denominator, making completion impossible.

Graphical Evaluation of Limits and Continuous Functions

• Limit Property of Continuous Functions:

  • If a function $f(x)$ is continuous at $x = a$, the limit as $x$ approaches $a$ equals the function value at $a$:         $\lim_{x \to a} f(x) = f(a)$
  • Graphical interpretation: Approaching $x = a$ using small intervals/arrows from both the left ($x < a$) and right ($x > a$) projects onto the graph toward the corresponding y-value $f(a)$.

• Example Limit Evaluations from a Continuous Graph:

  • At $x = 0$: Both left and right arrows on the graph point toward the origin $(0,0)$.         $\lim_{x \to 0} f(x) = f(0) = 0$
  • At $x = 1$: Graph projections from both directions point toward the peak at $(1,1)$.         $\lim_{x \to 1} f(x) = f(1) = 1$
  • At $x = 3$: Graph projections from both directions point toward the valley at $(3,-1)$.         $\lim_{x \to 3} f(x) = f(3) = -1$

Trigonometric Functions and Unit Circle Limits

• Unit Circle Definitions:

  • The unit circle is centered at the origin $(0,0)$ with radius $r = 1$, defined by equation:         $x^2 + y^2 = 1$
  • For an angle $x$ formed by a ray from the origin relative to the positive horizontal x-axis, the coordinates of the point on the unit circle are:         $(\cos(x), \sin(x))$

• First Quadrant Key Reference Values:

  • $\text{Angle } 0\text{\,rad}: (1, 0) \implies \cos(0) = 1, \sin(0) = 0$
  • $\text{Angle } \frac{\pi}{6}\text{\,rad}: \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \implies \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
  • $\text{Angle } \frac{\pi}{4}\text{\,rad}: \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \implies \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
  • $\text{Angle } \frac{\pi}{3}\text{\,rad}: \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \implies \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
  • $\text{Angle } \frac{\pi}{2}\text{\,rad}: (0, 1) \implies \cos\left(\frac{\pi}{2}\right) = 0, \sin\left(\frac{\pi}{2}\right) = 1$

• Mnemonic Device for First-Quadrant Values:

  • Note that $\sqrt{3} > 1$, which guarantees that $\frac{\sqrt{3}}{2} > \frac{1}{2}$.
  • At $\frac{\pi}{6}$, the reference triangle is wider than it is tall (longer x-base, shorter y-height). Thus, the x-coordinate must be the larger value:         $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
  • At $\frac{\pi}{3}$, the triangle is tilted vertically (shorter x-base, taller y-height). Thus, the y-coordinate must be the larger value:         $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \quad \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

• Evaluating Trigonometric Limits:

  • Since sine and cosine are continuous functions across all real numbers, limits are evaluated directly via function values.
  • Limit A:$\lim_{x \to 0} \sin(x) = \sin(0) = 0$
  • Limit B:$\lim_{x \to \frac{\pi}{2}} \sin(x) = \sin\left(\frac{\pi}{2}\right) = 1$
  • Limit C:$\lim_{x \to 0} \cos(x) = \cos(0) = 1$
  • Limit D:$\lim_{x \to \pi} \cos(x) = \cos(\pi) = -1$
    • Explanation: Angle $\pi$ corresponds to the leftmost point on the unit circle $(-1, 0)$; cosine is the x-coordinate.
  • Limit E:$\lim_{x \to \frac{\pi}{6}} \sin(x) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
  • Limit F:$\lim_{x \to \frac{2\pi}{3}} \cos(x) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$
    • Explanation: The supplementary angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$. Angle $\frac{2\pi}{3}$ lies in Quadrant II, where x-coordinates are negative. Reflecting the Quadrant I x-coordinate of $\frac{\pi}{3}$ gives $-\frac{1}{2}$.
  • Limit G:$\lim_{x \to -\frac{\pi}{3}} \cos(x) = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$
    • Explanation: Rotating clockwise by $\frac{\pi}{3}$ places the angle in Quadrant IV. In Quadrant IV, x-coordinates remain positive, so the cosine value remains $\frac{1}{2}$.
  • Limit H:$\lim_{x \to \frac{5\pi}{6}} \sin(x) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}$
    • Explanation: The supplementary angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$. Angle $\frac{5\pi}{6}$ lies in Quadrant II, where y-coordinates (sine) remain positive. Thus, sine matches $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

One-Sided Limits and Discontinuous Functions

• Definition and Concepts:

  • One-sided limits analyze function behavior when approaching a point strictly from values smaller than $a$ (left-hand) or strictly from values greater than $a$ (right-hand).
  • Breakdown into one-sided limits is essential when analyzing piecewise or discontinuous functions.

• Left-Handed Limit:

  • Notation:$\lim_{x \to a^-} f(x) = L$
  • The superscript minus sign $-$ on $a$ denotes approaching from the left.
  • Considers values of $x$ of the form $x = a - h$, where $h > 0$ is a small positive increment.

• Right-Handed Limit:

  • Notation:$\lim_{x \to a^+} f(x) = L$
  • The superscript plus sign $+$ on $a$ denotes approaching from the right.
  • Considers values of $x$ of the form $x = a + h$, where $h > 0$ is a small positive increment.

• Distinction Between Negative Signs in One-Sided Limit Notation:

  • In the expression $\lim_{x \to -1^-} g(x)$, the leading minus sign indicates the location on the x-axis ($x = -1$), whereas the superscript minus sign indicates the direction of approach (from the left).

• Evaluating One-Sided Limits on Discontinuous Function $g(x)$:

  • Left-Hand Limit at $x = -1$:
    • Approaching $x = -1$ from values less than $-1$ follows the graph segment pointing to point $(-1, -1)$.         $\lim_{x \to -1^-} g(x) = -1$
  • Right-Hand Limit at $x = -1$:
    • Approaching $x = -1$ from values greater than $-1$ follows the graph segment pointing to point $(-1, -2)$.         $\lim_{x \to -1^+} g(x) = -2$
  • Conclusion on Two-Sided Limits:
    • Since $\lim_{x \to -1^-} g(x) \neq \lim_{x \to -1^+} g(x)$, the two-sided limit $\lim_{x \to -1} g(x)$ does not exist.

Questions & Student Discussions

• Academic Requirements & AP Credit:
  • A score of $3$ on the AP Calculus test is insufficient to fulfill prerequisite credit requirements at certain universities.
  • Students retaking calculus must complete sequence coursework leading through Linear Algebra as well as Calculus-based Physics (Physics: Mechanics, and Electricity & Magnetism).
  • Computer Science concentrations such as Human-Centric Computer Science focus on computer graphics, visualization, and physics simulations, necessitating full physics and calculus sequences.