Calculus: Understanding Limits and Continuity

Introduction to Limits
  • Limit concept in calculus: As $x$ approaches a number, what value does $f(x)$ approach?
  • Example function: f(x)=x2−4x−2f(x) = \frac{x^2 - 4}{x - 2}
Applying Limits: Example with Direct Substitution
  • Evaluate lim⁡x→2f(x)\lim_{x \to 2} f(x)
  • Plugging in $x = 2$ gives:
    • Numerator: $2^2 - 4 = 0$
    • Denominator: $2 - 2 = 0$
    • Result: Undefined (indeterminate form: $ rac{0}{0}$)
Finding Limits Using Close Approximations
  • Use values close to 2 to find a limit:
    • Example with $x = 1.9$:
    • f(1.9)=(1.9)2−41.9−2=−0.39−0.1=3.9f(1.9) = \frac{(1.9)^2 - 4}{1.9 - 2} = \frac{-0.39}{-0.1} = 3.9
    • Example with $x = 1.99$:
    • f(1.99)=(1.99)2−41.99−2=−0.0399−0.01=3.99f(1.99) = \frac{(1.99)^2 - 4}{1.99 - 2} = \frac{-0.0399}{-0.01} = 3.99
  • As $x$ approaches 2, $f(x)$ approaches 4.
Simplifying Functions for Limits
  • Factor the numerator:
    • f(x)=(x+2)(x−2)x−2f(x) = \frac{(x + 2)(x - 2)}{x - 2}
    • Cancel $(x - 2)$ to find the new limit:
    • New expression: lim⁡x→2(x+2)\lim_{x \to 2} (x + 2)
    • Result after substitution: $2 + 2 = 4$.
Evaluating Further Limits - Problems
Problem 1
  • Limit as x→3x \to 3 of f(x)=x2+5x−4f(x) = x^2 + 5x - 4:
    • Direct substitution gives:
    • f(3)=32+5∗3−4=9+15−4=20f(3) = 3^2 + 5*3 - 4 = 9 + 15 - 4 = 20
Problem 2
  • Limit as x→3x \to 3 of f(x)=x2−8x+15x−3f(x) = \frac{x^2 - 8x + 15}{x - 3}:
    • Direct substitution undefined. Factor to find the limit:
    • x2−8x+15=(x−3)(x−5)x^2 - 8x + 15 = (x - 3)(x - 5)
    • Cancel $(x - 3)$:
    • lim⁡x→3(x−5)=3−5=−2\lim_{x \to 3} (x - 5) = 3 - 5 = -2
Problem 3
  • Complex fraction limit:
    • lim⁡x→41(x−1/4)/(x−4)\lim_{x \to 4} \frac{1}{(x - 1/4)}/(x - 4)
    • Multiply top/bottom by common denominator.
    • Resulting limit: lim⁡x→4−14x\lim_{x \to 4} \frac{-1}{4x}
    • After substitution: −116\frac{-1}{16}
Convergence to Limits via Approximations
  • Verify limits by plugging in close values:
    • Example: 4.1,4.014.1, 4.01 show convergence to $\frac{-1}{16}$
Limit Involving Square Roots
  • Problem: Find lim⁡x→9x−3x−9\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}:
  • Multiply by the conjugate x+3\sqrt{x} + 3.
  • Result after simplification yields final result: 16\frac{1}{6}.
Evaluating Limits Graphically

  • One-sided limits:

    • Example of approach to −3-3:
    • Left-hand limit approaches 1, right-hand limit approaches 2.

  • Conclusion: Limit does not exist since left and right-hand limits differ.

Types of Discontinuities
  • Jump Discontinuity: Occurs when limits from both sides differ.
  • Removable Discontinuity: Limit exists, but not equal to the function value.
  • Infinite Discontinuity: Approaches infinity.
  • Continuous Function: All limit values (one-sided and the function itself) match.
Additional Practice Problems

  1. Find limits approaching -1:

    • Left side: approaches -3.
    • Right side: approaches -3.
    • Function value $f(-1) = -2$, indicates a removable discontinuity.

  2. Find limits near -2:

    • Left side goes to +∞, right side goes to -∞, limit does not exist.

  3. Find limits approaching 1:

    • Both sides approach -1, with function value at $f(1) = -1$, indicating continuity.