Exhaustive Calculus Notes: Limits, Infinite Limits, and Algebraic Limit Calculations

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Introduction to Calculus and the Concept of Limits

  • Foundational Hierarchy of Calculus:

    • Calculus 1: Primarily focuses on derivatives.

    • Calculus 2 (or AP Calculus BC): Focuses on integrals, series, and integration applications.

    • The Foundational Concept: The limit is the underlying foundation of all calculus, as both derivatives and integrals are explicitly defined as limits.

  • Intuitive Definition of a Limit:

    • A limit examines the behavior of a function's yy-values as xx gets arbitrarily close to a specific target value aa.

    • Notation: \n\lim_{x \to a}\left(f(x)\right) = L\n

    • Meaning: As xx approaches aa from either side (without necessarily reaching aa), f(x)f(x) approaches LL.

    • Critical Principle: The limit as x→ax \to a depends strictly on values near aa. The actual value f(a)f(a) may be defined, undefined, or equal to a completely different value; f(a)f(a) has no impact on the limit as x→ax \to a.

  • Numerical Exploration Example:

    • Consider f(x)=xx+4−2f(x) = \frac{x}{\sqrt{x+4} - 2} near x=0x = 0.

    • Direct evaluation at x=0x = 0:     \n    f(0) = \frac{0}{\sqrt{0+4} - 2} = \frac{0}{2-2} = \frac{0}{0} \quad (\text{Undefined / Does Not Exist})\n    

    • Evaluating xx-values approaching 00 from the negative side (x<0x < 0):

    • x=−0.5  ⟹  y=3.87x = -0.5 \implies y = 3.87

    • x=−0.1  ⟹  y=3.97x = -0.1 \implies y = 3.97

    • x=−0.01  ⟹  y=3.98x = -0.01 \implies y = 3.98

    • x=−0.001  ⟹  y=3.99x = -0.001 \implies y = 3.99

    • Evaluating xx-values approaching 00 from the positive side (x>0x > 0):

    • x=0.5  ⟹  y=4.12x = 0.5 \implies y = 4.12

    • As x→0+x \to 0^+, values progressively decrease toward 44.

    • Conclusion: As x→0x \to 0, f(x)f(x) approaches 44. Thus, \n\lim_{x \to 0}\left(\frac{x}{\sqrt{x+4} - 2}\right) = 4\n

  • Graphical Evaluation Scenarios:

    • Scenario 1 (Hole at Target Point): f(x)f(x) has a hole at (4,2)(4, 2) with no point filled anywhere at x=4x = 4. Here, lim⁡x→4(f(x))=2\lim_{x \to 4}\left(f(x)\right) = 2, but f(4)f(4) is undefined.

    • Scenario 2 (Displaced Point): f(x)f(x) has a hole at (4,2)(4, 2) and a solid point at (4,3)(4, 3). Here, lim⁡x→4(f(x))=2\lim_{x \to 4}\left(f(x)\right) = 2, but f(4)=3f(4) = 3.

    • Scenario 3 (Continuous Point): f(x)f(x) passes continuously through (4,2)(4, 2) with a solid point. Here, lim⁡x→4(f(x))=2\lim_{x \to 4}\left(f(x)\right) = 2, and f(4)=2f(4) = 2.

One-Sided Limits and Graphical Evaluation

  • One-Sided Limit Definitions:

    • Left-Hand Limit (LHL): The behavior of f(x)f(x) as xx approaches aa purely from values strictly less than aa (the left side on the number line).     \n    \lim_{x \to a^-}\left(f(x)\right)\n    

    • Right-Hand Limit (RHL): The behavior of f(x)f(x) as xx approaches aa purely from values strictly greater than aa (the right side on the number line).     \n    \lim_{x \to a^+}\left(f(x)\right)\n    

  • Existence Criterion for Two-Sided Limits:

    • The general two-sided limit lim⁡x→a(f(x))=L\lim_{x \to a}\left(f(x)\right) = L exists if and only if both one-sided limits exist and are equal to the same value LL:     \n    \lim_{x \to a}\left(f(x)\right) = L \iff \lim_{x \to a^-}\left(f(x)\right) = L \quad \text{and} \quad \lim_{x \to a^+}\left(f(x)\right) = L\n    

    • If lim⁡x→a−(f(x))≠lim⁡x→a+(f(x))\lim_{x \to a^-}\left(f(x)\right) \neq \lim_{x \to a^+}\left(f(x)\right), then lim⁡x→a(f(x))=Does Not Exist (DNE)\lim_{x \to a}\left(f(x)\right) = \text{Does Not Exist (DNE)}.

  • Graphical Example Analysis:

    • Given a piecewise graph analyzed around x=4x = 4:

    • Approach from left: lim⁡x→4−(f(x))=2\lim_{x \to 4^-}\left(f(x)\right) = 2

    • Approach from right: lim⁡x→4+(f(x))=4\lim_{x \to 4^+}\left(f(x)\right) = 4

    • Overall limit: lim⁡x→4(f(x))=DNE\lim_{x \to 4}\left(f(x)\right) = \text{DNE} because 2≠42 \neq 4.

  • Comprehensive Graphical Analysis Examples:

    • For a piecewise graph evaluated at various xx-values:

    • At x=4x = 4:

      • lim⁡x→4−(f(x))=2\lim_{x \to 4^-}\left(f(x)\right) = 2

      • lim⁡x→4+(f(x))=2\lim_{x \to 4^+}\left(f(x)\right) = 2

      • lim⁡x→4(f(x))=2\lim_{x \to 4}\left(f(x)\right) = 2

    • At x=2x = 2:

      • lim⁡x→2−(f(x))=3\lim_{x \to 2^-}\left(f(x)\right) = 3

      • lim⁡x→2+(f(x))=6\lim_{x \to 2^+}\left(f(x)\right) = 6

      • lim⁡x→2(f(x))=DNE\lim_{x \to 2}\left(f(x)\right) = \text{DNE} (LHL ≠\neq RHL)

    • At x=1x = 1:

      • lim⁡x→1−(f(x))=−4\lim_{x \to 1^-}\left(f(x)\right) = -4

      • lim⁡x→1+(f(x))=−4\lim_{x \to 1^+}\left(f(x)\right) = -4

      • lim⁡x→1(f(x))=−4\lim_{x \to 1}\left(f(x)\right) = -4

    • At x=−6x = -6:

      • lim⁡x→−6−(f(x))=−1\lim_{x \to -6^-}\left(f(x)\right) = -1

      • lim⁡x→−6+(f(x))=7\lim_{x \to -6^+}\left(f(x)\right) = 7

      • lim⁡x→−6(f(x))=DNE\lim_{x \to -6}\left(f(x)\right) = \text{DNE} (LHL ≠\neq RHL)

    • At x=−3x = -3:

      • lim⁡x→−3−(f(x))=∞\lim_{x \to -3^-}\left(f(x)\right) = \infty

      • lim⁡x→−3+(f(x))=∞\lim_{x \to -3^+}\left(f(x)\right) = \infty

      • lim⁡x→−3(f(x))=∞\lim_{x \to -3}\left(f(x)\right) = \infty

    • At x=−1x = -1:

      • lim⁡x→−1(f(x))=−3\lim_{x \to -1}\left(f(x)\right) = -3

    • At x=6x = 6:

      • lim⁡x→6−(f(x))=−∞\lim_{x \to 6^-}\left(f(x)\right) = -\infty

      • lim⁡x→6+(f(x))=∞\lim_{x \to 6^+}\left(f(x)\right) = \infty

      • lim⁡x→6(f(x))=DNE\lim_{x \to 6}\left(f(x)\right) = \text{DNE}

Infinite Limits and Vertical Asymptotes

  • Understanding Infinite Limits:

    • If f(x)f(x) increases or decreases without bound as x→ax \to a, the limit is written as ∞\infty or −∞-\infty.

    • Infinity ( ∞ \,\infty\,) is not a real number; it describes the specific unbounded behavior of a function.

    • Reporting ∞\infty or −∞-\infty is a more descriptive and preferred answer than merely writing "DNE", though technically any limit equal to ±∞\pm\infty does not exist in terms of real numbers.

  • Basic Reciprocal Functions:

    • For f(x)=1x2f(x) = \frac{1}{x^2} as x→0x \to 0:

    • As xx gets closer to 00 from either side, x2x^2 becomes an extremely small positive number, making 1x2\frac{1}{x^2} extremely large.

    • lim⁡x→0(1x2)=∞\lim_{x \to 0}\left(\frac{1}{x^2}\right) = \infty

    • For f(x)=1xf(x) = \frac{1}{x} as x→0x \to 0:

    • lim⁡x→0+(1x)=∞\lim_{x \to 0^+}\left(\frac{1}{x}\right) = \infty

    • lim⁡x→0−(1x)=−∞\lim_{x \to 0^-}\left(\frac{1}{x}\right) = -\infty

    • lim⁡x→0(1x)=DNE\lim_{x \to 0}\left(\frac{1}{x}\right) = \text{DNE} because ∞≠−∞\infty \neq -\infty.

  • Calculus Definition of a Vertical Asymptote:

    • The line x=ax = a is a vertical asymptote of the function f(x)f(x) if at least one of the following six limit statements is true:

    • lim⁡x→a(f(x))=∞\lim_{x \to a}\left(f(x)\right) = \infty

    • lim⁡x→a−(f(x))=∞\lim_{x \to a^-}\left(f(x)\right) = \infty

    • lim⁡x→a+(f(x))=∞\lim_{x \to a^+}\left(f(x)\right) = \infty

    • lim⁡x→a(f(x))=−∞\lim_{x \to a}\left(f(x)\right) = -\infty

    • lim⁡x→a−(f(x))=−∞\lim_{x \to a^-}\left(f(x)\right) = -\infty

    • lim⁡x→a+(f(x))=−∞\lim_{x \to a^+}\left(f(x)\right) = -\infty

    • Example: From the graphical analysis above, the graph exhibits vertical asymptotes at x=−3x = -3 and x=6x = 6.

Evaluating Infinite Limits Algebraically (Non-Zero / Zero)

  • The Non-Zero0\frac{\text{Non-Zero}}{0} Form Rule:

    • Direct substitution yielding c0\frac{c}{0} (where c≠0c \neq 0) indicates the presence of a vertical asymptote.

    • The result of such a limit is guaranteed to be either ∞\infty or −∞-\infty (or DNE if a two-sided limit approaches opposite infinities).

    • Methodology (Test Point Sign Analysis):

    1. Perform direct substitution to confirm the nonzero0\frac{\text{nonzero}}{0} format.

    2. Select a test point extremely close to aa on the designated side (e.g., a−0.1a - 0.1 for left-hand limits, a+0.1a + 0.1 for right-hand limits).

    3. Evaluate the sign (++ or −-) of the numerator and denominator to determine if the quotient approaches positive or negative infinity.

  • Worked Examples:

    • Example 1: lim⁡x→−5−(x−1x+5)\lim_{x \to -5^-}\left(\frac{x-1}{x+5}\right)

    • Direct Substitution: −5−1−5+5=−60\frac{-5-1}{-5+5} = \frac{-6}{0}

    • Test Point: Choose x=−5.1x = -5.1 (left of −5-5):

      • Numerator: −5.1−1=−6.1-5.1 - 1 = -6.1 (Negative)

      • Denominator: −5.1+5=−0.1-5.1 + 5 = -0.1 (Negative)

      • Quotient Sign: NegativeNegative=Positive\frac{\text{Negative}}{\text{Negative}} = \text{Positive}

    • Result: lim⁡x→−5−(x−1x+5)=∞\lim_{x \to -5^-}\left(\frac{x-1}{x+5}\right) = \infty

    • Example 2: lim⁡x→−5+(x−1x+5)\lim_{x \to -5^+}\left(\frac{x-1}{x+5}\right)

    • Direct Substitution: −60\frac{-6}{0}

    • Test Point: Choose x=−4.9x = -4.9 (right of −5-5):

      • Numerator: −4.9−1=−5.9-4.9 - 1 = -5.9 (Negative)

      • Denominator: −4.9+5=+0.1-4.9 + 5 = +0.1 (Positive)

      • Quotient Sign: NegativePositive=Negative\frac{\text{Negative}}{\text{Positive}} = \text{Negative}

    • Result: lim⁡x→−5+(x−1x+5)=−∞\lim_{x \to -5^+}\left(\frac{x-1}{x+5}\right) = -\infty

    • Example 3: lim⁡x→−5(x−1x+5)\lim_{x \to -5}\left(\frac{x-1}{x+5}\right)

    • Since lim⁡x→−5−(x−1x+5)=∞\lim_{x \to -5^-}\left(\frac{x-1}{x+5}\right) = \infty and lim⁡x→−5+(x−1x+5)=−∞\lim_{x \to -5^+}\left(\frac{x-1}{x+5}\right) = -\infty, the two-sided limit Does Not Exist (DNE) because LHL ≠\neq RHL.

    • Example 4: lim⁡x→2−(x+32−x)\lim_{x \to 2^-}\left(\frac{x+3}{2-x}\right)

    • Direct Substitution: 2+32−2=50\frac{2+3}{2-2} = \frac{5}{0}

    • Test Point: Choose x=1.9x = 1.9 (left of 22):

      • Denominator: 2−1.9=+0.12 - 1.9 = +0.1 (Positive)

      • Quotient Sign: PositivePositive=Positive\frac{\text{Positive}}{\text{Positive}} = \text{Positive}

    • Result: lim⁡x→2−(x+32−x)=∞\lim_{x \to 2^-}\left(\frac{x+3}{2-x}\right) = \infty

    • Example 5: lim⁡x→3(x3(x−6)(x−3)2)\lim_{x \to 3}\left(\frac{x^3}{(x-6)(x-3)^2}\right)

    • Direct Substitution: 33(3−6)(3−3)2=27−3⋅(0)2=270\frac{3^3}{(3-6)(3-3)^2} = \frac{27}{-3 \cdot (0)^2} = \frac{27}{0}

    • Structural Sign Analysis without explicit test points:

      • Numerator: 33=273^3 = 27 (Positive)

      • Factor (x−6)(x-6): 3−6=−33 - 6 = -3 (Negative)

      • Factor (x−3)2(x-3)^2: Any non-zero value squared is strictly positive (>0> 0) from both the left and right sides.

      • Denominator Sign: (Negative)⋅(Positive)=Negative(\text{Negative}) \cdot (\text{Positive}) = \text{Negative}

      • Overall Sign: PositiveNegative=Negative\frac{\text{Positive}}{\text{Negative}} = \text{Negative}

    • Result: lim⁡x→3(x3(x−6)(x−3)2)=−∞\lim_{x \to 3}\left(\frac{x^3}{(x-6)(x-3)^2}\right) = -\infty

Logarithmic Limits

  • Behavior of the Natural Logarithm near Zero:

    • The function f(x)=ln⁡(x)f(x) = \ln(x) is defined strictly for x>0x > 0.

    • Fundamental Limit: \n\lim_{x \to 0^+}\left(\ln(x)\right) = -\infty\n

    • Left-hand limit lim⁡x→0−(ln⁡(x))\lim_{x \to 0^-}\left(\ln(x)\right) does not exist because negative numbers are outside the domain.

  • Composite Logarithmic Limit Example:

    • Evaluate lim⁡x→4+(ln⁡(x3−64))\lim_{x \to 4^+}\left(\ln(x^3 - 64)\right)

    • Direct evaluation of inner expression at x=4x = 4: 43−64=04^3 - 64 = 0.

    • Test point slightly to the right (x=4.1x = 4.1): 4.13−64>04.1^3 - 64 > 0 (approaches 00 through positive values, 0+0^+).

    • Applying logarithmic behavior: As inner argument u→0+u \to 0^+, ln⁡(u)→−∞\ln(u) \to -\infty.

    • Result: lim⁡x→4+(ln⁡(x3−64))=−∞\lim_{x \to 4^+}\left(\ln(x^3 - 64)\right) = -\infty

Identifying Vertical Asymptotes vs. Holes

  • Rational Functions Candidate Analysis:

    • Vertical asymptotes occur where the denominator equals 00, provided the numerator does not simultaneously equal 00 (or provided the limit evaluates to nonzero0\frac{\text{nonzero}}{0}). If both numerator and denominator equal 00, the point represents a removable discontinuity (a hole) rather than a vertical asymptote.

  • Comparative Candidate Example:

    • Function: f(x)=x−2(x−2)(x−4)f(x) = \frac{x-2}{(x-2)(x-4)}

    • Setting denominator to zero yields candidates: x=2x = 2 and x=4x = 4.

    • Testing x=2x = 2:

    • Direct Substitution: 2−2(2−2)(2−4)=00\frac{2-2}{(2-2)(2-4)} = \frac{0}{0}

    • Does not match nonzero0\frac{\text{nonzero}}{0} format; x=2x = 2 is a hole (removable discontinuity), not a vertical asymptote.

    • Testing x=4x = 4:

    • Direct Substitution: 4−2(4−2)(4−4)=20\frac{4-2}{(4-2)(4-4)} = \frac{2}{0}

    • Matches nonzero0\frac{\text{nonzero}}{0} format; x=4x = 4 is a confirmed vertical asymptote.

Calculating Limits Algebraically (Section 2.3)

  • Fundamental Limit Laws (assuming lim⁡x→a(f(x))\lim_{x \to a}\left(f(x)\right) and lim⁡x→a(g(x))\lim_{x \to a}\left(g(x)\right) exist):

    1. Sum/Difference Rule: lim⁡x→a(f(x)±g(x))=lim⁡x→a(f(x))±lim⁡x→a(g(x))\lim_{x \to a}\left(f(x) \pm g(x)\right) = \lim_{x \to a}\left(f(x)\right) \pm \lim_{x \to a}\left(g(x)\right)

    2. Constant Multiple Rule: lim⁡x→a(c⋅f(x))=c⋅lim⁡x→a(f(x))\lim_{x \to a}\left(c \cdot f(x)\right) = c \cdot \lim_{x \to a}\left(f(x)\right)

    3. Product Rule: lim⁡x→a(f(x)⋅g(x))=lim⁡x→a(f(x))⋅lim⁡x→a(g(x))\lim_{x \to a}\left(f(x) \cdot g(x)\right) = \lim_{x \to a}\left(f(x)\right) \cdot \lim_{x \to a}\left(g(x)\right)

    4. Quotient Rule: lim⁡x→a(f(x)g(x))=lim⁡x→a(f(x))lim⁡x→a(g(x))\lim_{x \to a}\left(\frac{f(x)}{g(x)}\right) = \frac{\lim_{x \to a}\left(f(x)\right)}{\lim_{x \to a}\left(g(x)\right)} provided lim⁡x→a(g(x))≠0\lim_{x \to a}\left(g(x)\right) \neq 0

    5. Power Rule: lim⁡x→a([f(x)]n)=[lim⁡x→a(f(x))]n\lim_{x \to a}\left([f(x)]^n\right) = \left[\lim_{x \to a}\left(f(x)\right)\right]^n

  • Rigorous Limit Law Step-by-Step Proof Demonstration:

    • Evaluate lim⁡x→−2((3x2+5x+1)4)\lim_{x \to -2}\left((3x^2 + 5x + 1)^4\right) using formal limit rules:     \n    \begin{aligned}\n    \lim_{x \to -2}\left((3x^2 + 5x + 1)^4\right) &= \left[\lim_{x \to -2}\left(3x^2 + 5x + 1\right)\right]^4 \quad &(\text{Rule 5: Power Rule}) \\\n    &= \left[\lim_{x \to -2}\left(3x^2\right) + \lim_{x \to -2}\left(5x\right) + \lim_{x \to -2}\left(1\right)\right]^4 \quad &(\text{Rule 1: Sum Rule}) \\\n    &= \left[3 \lim_{x \to -2}\left(x^2\right) + 5 \lim_{x \to -2}\left(x\right) + \lim_{x \to -2}\left(1\right)\right]^4 \quad &(\text{Rule 2: Constant Multiple}) \\\n    &= \left[3(-2)^2 + 5(-2) + 1\right]^4 \\\n    &= [12 - 10 + 1]^4 = 3^4 = 81\n    \end{aligned}\n    

  • Direct Substitution Property:

    • If ff is a polynomial or rational function and aa is in the domain of ff, then lim⁡x→a(f(x))=f(a)\lim_{x \to a}\left(f(x)\right) = f(a).

    • Full limit law expansions are shown for theoretical justification, but direct substitution is used in practice whenever defined.

  • Limit Law Given Abstract Given Values:

    • Given lim⁡x→3(f(x))=16\lim_{x \to 3}\left(f(x)\right) = 16, evaluate:     \n    \lim_{x \to 3}\left(\frac{\sqrt{x^2 - 4} \cdot f(x)}{f(x) + x + 2}\right)\n    

    • Apply limit properties to evaluate components individually at x=3x = 3:     \n    \frac{\sqrt{3^2 - 4} \cdot \lim_{x \to 3}\left(f(x)\right)}{\lim_{x \to 3}\left(f(x)\right) + 3 + 2} = \frac{\sqrt{5} \cdot 16}{16 + 3 + 2} = \frac{16\sqrt{5}}{21}\n    

    • Alternative equivalent form: 8021\frac{\sqrt{80}}{21}.

Indeterminate Forms (0/0) and Algebraic Strategies

  • Definition of Indeterminate Form 00\frac{0}{0}:

    • When direct substitution into a limit yields 00\frac{0}{0}, the result cannot be determined by inspection.

    • Mathematical conflict: The numerator approaching 00 pulls the fraction toward 00, while the denominator approaching 00 pulls the fraction toward ±∞\pm\infty. The algebraic structure determines which behavior prevails.

  • Strict Rule on L'Hôpital's Rule:

    • L'Hôpital's Rule MUST NOT be used on written workout exams prior to its official introduction in the curriculum.

    • Logical reasoning: Derivatives are defined using limits. Using derivative rules (L'Hôpital) to evaluate limits creates circular logic.

  • Strict Notation Requirements:

    • The limit operator lim⁡x→a\lim_{x \to a} must be explicitly written at every single step of algebraic simplification until direct substitution is performed.

  • Algebraic Strategy 1: Factoring and Canceling:

    • Evaluate lim⁡x→4(x2−x−12x2−16)\lim_{x \to 4}\left(\frac{x^2 - x - 12}{x^2 - 16}\right)

    • Step 1: Direct Substitution   ⟹  42−4−1242−16=00\implies \frac{4^2 - 4 - 12}{4^2 - 16} = \frac{0}{0} (Indeterminate Form).

    • Step 2: Factor both numerator and denominator:       \n      \lim_{x \to 4}\left(\frac{(x-4)(x+3)}{(x-4)(x+4)}\right)\n      

    • Step 3: Cancel common factor (x−4)(x-4). This cancellation is valid under limits because limits evaluate behavior where x≠4x \neq 4.       \n      \lim_{x \to 4}\left(\frac{x+3}{x+4}\right)\n      

    • Step 4: Direct Substitution:       \n      \frac{4+3}{4+4} = \frac{7}{8}\n      

  • Mathematical Nuance on Function Equality:

    • The functions g(x)=(x−4)(x+3)(x−4)(x+4)g(x) = \frac{(x-4)(x+3)}{(x-4)(x+4)} and h(x)=x+3x+4h(x) = \frac{x+3}{x+4} are not strictly identical functions because g(4)g(4) is undefined while h(4)=78h(4) = \frac{7}{8}.

    • They are equal for all x≠4x \neq 4. Because limits specifically care about x→4x \to 4 (x≠4x \neq 4), replacing g(x)g(x) with h(x)h(x) inside the limit is mathematically rigorous.

  • Algebraic Strategy 2: Expanding and Factoring:

    • Evaluate lim⁡h→0((h−4)2−16h)\lim_{h \to 0}\left(\frac{(h-4)^2 - 16}{h}\right)

    • Step 1: Direct Substitution   ⟹  (0−4)2−160=16−160=00\implies \frac{(0-4)^2 - 16}{0} = \frac{16-16}{0} = \frac{0}{0} (Indeterminate Form).

    • Step 2: Expand the squared binomial numerator:       \n      \lim_{h \to 0}\left(\frac{h^2 - 8h + 16 - 16}{h}\right) = \lim_{h \to 0}\left(\frac{h^2 - 8h}{h}\right)\n      

    • Step 3: Factor hh out of the numerator (recommended step to prevent cancellation errors):       \n      \lim_{h \to 0}\left(\frac{h(h - 8)}{h}\right)\n      

    • Step 4: Cancel hh:       \n      \lim_{h \to 0}\left(h - 8\right)\n      

    • Step 5: Direct Substitution:       \n      0 - 8 = -8\n      

Dialogue and Student Interactions

  • Student Clarification on Graph Reading:

    • Student Question: On reading the graph around x=1x = 1, a student noted that the graph level appeared closer to y=−4y = -4 rather than y=−3y = -3.

    • Instructor Response: The instructor acknowledged that graph labels could be clearer and confirmed that y=−4y = -4 is the intended precise reading for that point on the graph.

  • Student Interactions Post-Lecture:

    • Students discussed post-lecture plans, engineering coursework, dining hall schedules (eating at Zachry/Zachs at 12:10 PM), off-campus housing (North Plain Crossing), resume preparation for upcoming engineering events, and high school AP Chemistry credit exemptions.