Calculus: Limits, Derivatives, and Integration

Limits

• Purpose of a limit

  • Lets us predict the value a function "wants" to take as $x$ approaches a specific point, even when the function is undefined at that point.

  • Provides insight into a function’s behaviour "near" rather than "at" the point.

• Foundational example

  • Function: $f(x)=\dfrac{x^2-4}{x-2}$

  • Direct substitution $x=2$ gives $\tfrac{0}{0}$ (indeterminate), so $f(2)$ is undefined.

  • Numerical approach

  • $x=2.1 \Rightarrow f(2.1) \approx 4.01$

  • $x=2.01 \Rightarrow f(2.01) \approx 4.0001$

  • Trend → function values approach $4$.

  • Algebraic confirmation

  • Factor numerator (difference of squares):
    $x^2-4=(x+2)(x-2)$

  • Cancel common factor with denominator:
    $f(x)=\dfrac{(x+2)(x-2)}{x-2}=x+2\quad (x\neq2)$

  • Limit statement:
    $\displaystyle \lim_{x\to2}\dfrac{x^2-4}{x-2}=2+2=4$

  • Key takeaway: the limit exists (equals $4$) even though $f(2)$ does not.

• Conceptual role of limits

  • Underpin both derivatives (instantaneous rate) and integrals (accumulated area).

  • Bridge between discrete approximations and exact values.

Derivatives

• Definition & intuition

  • A derivative is a function giving the slope of the tangent line to the original function at every $x$.

  • Describes instantaneous rate of change.

• Power-rule example

  • Original function: $f(x)=x^4$

  • Derivative: $f'(x)=4x^{4-1}=4x^3$

• Geometric meaning via $f(x)=x^3$

  • Goal: slope of tangent at $x=2$.

  • Direct derivative:
    $f'(x)=3x^2 \;\Rightarrow\; f'(2)=3\cdot2^2=12$

  • Interpretation: move $1$ unit right → $y$ increases $12$ units.

• Secant-line approximations (numerical link to limits)

  1. Wide interval $[1,3]$

  • $f(3)=27,\;f(1)=1$

  • $m=\dfrac{27-1}{3-1}=13$

  1. Narrow interval $[1.9,2.1]$

  • $f(2.1)=2.1^3\approx9.261\;\text{and}\;f(1.9)=1.9^3\approx6.859$

  • $m=\dfrac{9.261-6.859}{2.1-1.9}\approx12.01$

    • As endpoints approach $2$, slope → $12$, matching the derivative.

• Limit definition (formal tie-in)
  $\displaystyle f'(2)=\lim_{x\to2}\frac{f(x)-f(2)}{x-2}$

  • For $f(x)=x^3$, rewrite numerator as a difference of cubes:
    $x^3-8=(x-2)(x^2+2x+4)$

  • Cancelling $x-2$ yields limit value $12$.

• Multiple computation paths

  • Direct differentiation rules (fast).

  • Secant-line limits (conceptual foundation).

Integration (Antiderivatives)

• Core idea

  • Reverse process of differentiation; recovers accumulated quantity.

  • Often phrased as “area under a curve,” but applies to any accumulation.

• Basic antiderivative example

  • Given $f'(x)=4x^3$, integrate to get original:
    $\int4x^3\,dx=4\cdot\frac{x^{3+1}}{3+1}+C=x^4+C$

  • Constant $C$ captures any vertical shift lost during differentiation.

• Summary comparison: Derivative vs. Integral

  • Derivative → instant rate ($\frac{dy}{dx}$ ‑ a division perspective).

  • Integral → total accumulation (area ≈ $y\times x$ ‑ a multiplication perspective).

  • They are inverse operations:
    $\frac{d}{dx}\bigl(\int f(x)\,dx\bigr)=f(x)\quad\text{and}\quad\int f'(x)\,dx=f(x)+C.$

Application 1: Volume-in-Tank Function

• Stored-water model
  $A(t)=0.01t^{2}+0.5t+100$

  • $t$ in minutes; $A(t)$ in gallons.

• Tabulated values

  • $A(0)=100$

  • $A(9)=105.31$

  • $A(10)=106$

  • $A(11)=106.71$

  • $A(20)=140$

• Instantaneous rate at $t=10$

  • Derivative: $A'(t)=0.02t+0.5$

  • Evaluate: $A'(10)=0.02(10)+0.5=0.2+0.5=0.7$

  • Interpretation: water level rising $0.7\,\text{gal}\,/\,\text{min}$ at that instant.

  • Check with average differences: between 9→10 (≈0.69) and 10→11 (≈0.71) gallons → derivative (~0.70) matches central trend.

Application 2: Inflow-Rate Function (Definite Integral)

• Flow-rate model
  $r(t)=0.5t+20$ (gallons per minute)

• Desired accumulation: $t=20 \text{ to } t=100$

  • Use definite integral:
    $\displaystyle\Delta V=\int_{20}^{100}(0.5t+20)\,dt$

• Antiderivative & evaluation
  $\int(0.5t+20)\,dt=0.5\,\frac{t^{2}}{2}+20t=0.25t^{2}+20t$

  • Plug in limits:
    \begin{aligned}
    V(100)-V(20)
    &= \bigl(0.25\cdot100^{2}+20\cdot100\bigr)
    -\bigl(0.25\cdot20^{2}+20\cdot20\bigr)\
    &= \bigl(0.25\cdot10{,}000+2{,}000\bigr)
    -\bigl(0.25\cdot400+400\bigr)\
    &= \bigl(2{,}500+2{,}000\bigr)
    -\bigl(100+400\bigr)=4{,}500-500=4{,}000\;\text{gallons}
    \end{aligned}

• Graphical perspective

  • Plot $r(t)$; the shaded area from $t=20$ to $t=100$ equals $4{,}000$ gallons.

  • Represents change in volume, not total in tank at $t=100$.

Key Takeaways

• Limits: tool for analyzing near-point behaviour; formal foundation for derivatives and integrals.

• Derivatives: give slope/instantaneous rate; computed via rules or limit definition; practical for velocity, growth rates, etc.

• Integration: accumulates quantity over interval; inverse of differentiation; practical for total distance, volume, mass, etc.

• Real-world problems often pair both ideas: rate given → integrate to accumulate; accumulated amount given → differentiate for rate.

• Understanding their relationship (Fundamental Theorem of Calculus) unifies the two halves of calculus.