Introduction to Limits and Limit Properties

Introduction to Limits and Notation

  • Definitions and Fundamental Concepts:
    • A limit is defined as a yy value approached by a function as the xx value used in the function approaches a fixed value of cc.
    • The fundamental principle of a limit is that it represents an approached value, not necessarily the value of the function at that exact point.
  • Notation:
    • Standard notation for a limit uses a script abbreviation: lim⁡x→cf(x)\lim_{x \to c} f(x).
    • This expression is read as "the limit as xx approaches cc of f(x)f(x)."
  • Conditions for Existence:
    • For a limit to exist, the function must approach the same yy value from both the left side and the right side of the constant cc.
    • If the paths from either side do not meet at the same location, the limit does not exist (DNE).

Graphical Analysis of Limits

  • Function Verification:
    • Before evaluating limits graphically, verify the relation is a function using the vertical line test. A relation is a function if every vertical line intersects the graph no more than once.
  • Comparative Analysis of Limits and Function Values:
    • Example Case 1: At x=−6x = -6, the function approaches the same location from both sides. The limit is 33.
    • Example Case 2: At x=−3x = -3, the paths from the left and right do not meet. Consequently, lim⁡x→−3f(x)\lim_{x \to -3} f(x) does not exist. However, the actual function value exists, defined as f(−3)=1f(-3) = 1.
    • Example Case 3: At x=0x = 0, both sides of the graph converge at a single point on the yy-axis, yielding a limit of −2-2.
    • Example Case 4: At x=2x = 2, there is a hole in the graph. The limit as xx approaches 22 is −1-1, even though the point is not there. The actual function value is located elsewhere, such as at a solid dot at f(2)=3f(2) = 3.
    • Example Case 5: At x=5x = 5, the sides approach different values (the left at 22 and the right at 55). In this case, the limit does not exist.
  • Common Reasons for Non-Existence:
    • The primary reason a limit fails to exist is that the two sides of the function do not approach the same value.
    • Specificity can be added by noting that one side shoots toward positive infinity while the other shoots toward negative infinity, or that they simply approach distinct finite numbers.

Infinite Limits and Asymptotes

  • Vertical Asymptotes:
    • Consider the function g(x)=1x2g(x) = \frac{1}{x^2}. As xx approaches 00, the graph shoots upward infinitely on both sides.
    • While some perspectives classify this as "does not exist" due to the lack of a finite value, it is standard to identify the limit as ∞\infty when both sides approach that same direction within the context of calculus.
    • For the reflected function h(x)=−1x2h(x) = -\frac{1}{x^2}, as xx approaches 00, both sides shoot downward, meaning the limit is −infinity-\text{infinity}.

Numerical Estimation via Tabular Approximation

  • Methodology:
    • When a graph is unavailable and algebraic manipulation is unknown, a tabular or table approximation can be used to estimate a limit.
    • This involves selecting xx values increasingly close to the target value cc from both the left and the right sides.
  • Example: Estimating the limit as xx approaches 22:
    • From the left:
      • At x=1.8x = 1.8, f(x) is 0.67214f(x) \text{ is } 0.67214.
      • At x=1.9x = 1.9, f(x) is 0.57216f(x) \text{ is } 0.57216.
      • At x=1.99x = 1.99, f(x) is 0.50633f(x) \text{ is } 0.50633.
      • At x=1.999x = 1.999, f(x) is 0.50063f(x) \text{ is } 0.50063.
    • From the right:
      • At x=2.2x = 2.2, f(x) is 0.40056f(x) \text{ is } 0.40056.
      • At x=2.1x = 2.1, f(x) is 0.4446f(x) \text{ is } 0.4446.
      • At x=2.01x = 2.01, f(x) is 0.49381f(x) \text{ is } 0.49381.
      • At x=2.001x = 2.001, f(x) is 0.49938f(x) \text{ is } 0.49938.
    • Convergence: Since both sides appear to converge toward a specific decimal, the estimated limit is approximately 0.50.5.
  • Example: Estimating the limit as xx approaches −1-1 for x3+15x+5\frac{x^3+1}{5x+5}:
    • Using values like −1.2-1.2, −1.1-1.1, and −1.01-1.01, the results show an approach toward values around 1.661.66 or 1.671.67.

The Formal Epsilon-Delta Definition of a Limit

  • Definition Statement:
    • Let LL be a finite number, and let f(x)f(x) be defined on an open interval containing the value cc.
    • The statement lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L means that for any ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that if ∣x−c∣<ρ|x - c| < \rho, then ∣f(x)−L∣<ρ|f(x) - L| < \rho.
  • Conceptual Meaning:
    • ϵ\epsilon (epsilon) represents a chosen vertical distance or interval around the limit value LL.
    • δ\delta (delta) represents the required horizontal distance or interval around the target value cc needed to guarantee that every function value falls within the ϵ\epsilon interval of LL.
    • If one can specify a δ\delta for any given ϵ\epsilon, the limit is proven to be LL.
  • Computational Examples for δ\delta:
    • Linear Case: Given f(x)=6x+4f(x) = 6x + 4 and c=5c = 5, find the required δ\delta for ϵ=0.001\epsilon = 0.001.
      1. Determine LL by evaluating f(5)=6(5)+4=34f(5) = 6(5) + 4 = 34.
      2. Set up the inequality ∣f(x)−L∣<0.001|f(x) - L| < 0.001, leading to ∣6x+4−34∣<0.001|6x + 4 - 34| < 0.001.
      3. Simplify to ∣6x−30∣<0.001|6x - 30| < 0.001, then factor out the constant to get 6∣x−5∣<0.0016|x - 5| < 0.001.
      4. Divide by the constant to isolate the x−cx-c form: ∣x−5∣<0.0016|x - 5| < \frac{0.001}{6}.
      5. Therefore, δ=0.0016\delta = \frac{0.001}{6}.
    • Variable Case: Given f(x)=8−13xf(x) = 8 - \frac{1}{3}x and c=24c = 24, find δ\delta in terms of ϵ\epsilon where L=0L = 0.
      1. Set up ∣8−13x−0∣<ρ|8 - \frac{1}{3}x - 0| < \rho.
      2. Factor out the coefficient to isolate xx: ∣−13(x−24)∣<ρ|-\frac{1}{3}(x - 24)| < \rho.
      3. Move the scalar outside the absolute value sign: 13∣x−24∣<ρ\frac{1}{3}|x - 24| < \rho.
      4. Multiply by 33 to find the delta form: ∣x−24∣<3ρ|x - 24| < 3\rho.
      5. Therefore, δ=3ρ\delta = 3\rho.

Algebraic Properties of Limits

  • Foundational Laws (where kk is a constant):
    1. Constant Law: lim⁡x→ck=k\lim_{x \to c} k = k.
    2. Identity Law: lim⁡x→cx=c\lim_{x \to c} x = c.
    3. Scalar Multiple Law: lim⁡x→c[k×f(x)]=k×limx→cf(x)\lim_{x \to c} [k \times f(x)] = k \times \text{lim}_{x \to c} f(x).
    4. Sum/Difference Law: lim⁡x→c[f(x) plus or minus g(x)]=limx→cf(x) plus or minus limx→cg(x)\lim_{x \to c} [f(x) \text{ plus or minus } g(x)] = \text{lim}_{x \to c} f(x) \text{ plus or minus } \text{lim}_{x \to c} g(x).
    5. Product Law: lim⁡x→c[f(x)×g(x)]=[limx→cf(x)]×[limx→cg(x)]\lim_{x \to c} [f(x) \times g(x)] = [\text{lim}_{x \to c} f(x)] \times [\text{lim}_{x \to c} g(x)].
    6. Quotient Law: lim⁡x→cf(x)g(x)=limx→cf(x)limx→cg(x)\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\text{lim}_{x \to c} f(x)}{\text{lim}_{x \to c} g(x)}, provided the denominator limit is non-zero.
    7. Power Law: lim⁡x→c[f(x)]n=[limx→cf(x)]n\lim_{x \to c} [f(x)]^n = [\text{lim}_{x \to c} f(x)]^n for integer powers.
    8. Composition Law: lim⁡x→cf(g(x))=f(limx→cg(x))\lim_{x \to c} f(g(x)) = f(\text{lim}_{x \to c} g(x)).

Evaluation Examples Using Limit Properties

  • Multi-Step Evaluation:
    • Given lim⁡x→4f(x)=−3\lim_{x \to 4} f(x) = -3 and lim⁡x→4g(x)=5\lim_{x \to 4} g(x) = 5, evaluate lim⁡x→4[6f(x)+g(x)2+x]\lim_{x \to 4} [6 f(x) + g(x)^2 + x].
      1. Distribute limits over the sums: lim⁡6f(x)+lim g(x)2+lim x\lim 6 f(x) + \text{lim } g(x)^2 + \text{lim } x.
      2. Pull constants and use power rules: 6×(lim f(x))+(lim g(x))2+lim x6 \times (\text{lim } f(x)) + (\text{lim } g(x))^2 + \text{lim } x.
      3. Substitute known values: 6(−3)+(5)2+46(-3) + (5)^2 + 4.
      4. Calculation: −18+25+4=11-18 + 25 + 4 = 11.
  • Rational Function Evaluation:
    • Evaluate lim⁡x→42g(x)+13f(x)g(x)−x\lim_{x \to 4} \frac{2g(x) + 13}{f(x)g(x) - x}.
      1. Apply the quotient rule to separate top and bottom limits.
      2. Apply the scalar and product rules: 2×lim g(x)+13(lim f(x)×lim g(x))−lim x\frac{2 \times \text{lim } g(x) + 13}{(\text{lim } f(x) \times \text{lim } g(x)) - \text{lim } x}.
      3. Substitute values: 2(5)+13(−3)(5)−4\frac{2(5) + 13}{(-3)(5) - 4}.
      4. Calculation: 23−19\frac{23}{-19}.
  • Evaluation via Direct Substitution:
    • For continuous functions, properties allow for the direct substitution of the approach value cc for every occurrence of xx.
    • Polynomial example: lim⁡x→−5(7x−x3)=7(−5)−(−5)3=−35+125=90\lim_{x \to -5} (7x - x^3) = 7(-5) - (-5)^3 = -35 + 125 = 90.
    • Square root example: lim⁡x→−3sqrt(x2+12x+29)=sqrt((−3)2+12(−3)+29)=sqrt(9−36+29)=sqrt(2)\lim_{x \to -3} \text{sqrt}(x^2 + 12x + 29) = \text{sqrt}((-3)^2 + 12(-3) + 29) = \text{sqrt}(9 - 36 + 29) = \text{sqrt}(2).
    • Trigonometric example: lim⁡x→pi6sin(2x)=sin(2×pi6)=sin(pi3)=sqrt(3)2\lim_{x \to \frac{\text{pi}}{6}} \text{sin}(2x) = \text{sin}(2 \times \frac{\text{pi}}{6}) = \text{sin}(\frac{\text{pi}}{3}) = \frac{\text{sqrt}(3)}{2}.
    • Natural log example: Evaluation of a function results in the simplified form ln⁡(104−100)=ln(4)\ln(104 - 100) = \text{ln}(4).