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: $f(x) = \frac{x^2 - 4}{x - 2}$

Applying Limits: Example with Direct Substitution

• Evaluate $\lim_{x \to 2} f(x)$
• 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 $x = 1.9$:
  • $f(1.9) = \frac{(1.9)^2 - 4}{1.9 - 2} = \frac{-0.39}{-0.1} = 3.9$
  • Example with $x = 1.99$:
  • $f(1.99) = \frac{(1.99)^2 - 4}{1.99 - 2} = \frac{-0.0399}{-0.01} = 3.99$
• As $x$ approaches 2, $f(x)$ approaches 4.

Simplifying Functions for Limits

• Factor the numerator:
  • $f(x) = \frac{(x + 2)(x - 2)}{x - 2}$
  • Cancel $(x - 2)$ to find the new limit:
  • New expression: $\lim_{x \to 2} (x + 2)$
  • Result after substitution: $2 + 2 = 4$.

Evaluating Further Limits - Problems

Problem 1

• Limit as $x \to 3$ of $f(x) = x^2 + 5x - 4$:
  • Direct substitution gives:
  • $f(3) = 3^2 + 5*3 - 4 = 9 + 15 - 4 = 20$

Problem 2

• Limit as $x \to 3$ of $f(x) = \frac{x^2 - 8x + 15}{x - 3}$:
  • Direct substitution undefined. Factor to find the limit:
  • $x^2 - 8x + 15 = (x - 3)(x - 5)$
  • Cancel $(x - 3)$:
  • $\lim_{x \to 3} (x - 5) = 3 - 5 = -2$

Problem 3

• Complex fraction limit:
  • $\lim_{x \to 4} \frac{1}{(x - 1/4)}/(x - 4)$
  • Multiply top/bottom by common denominator.
  • Resulting limit: $\lim_{x \to 4} \frac{-1}{4x}$
  • After substitution: $\frac{-1}{16}$

Convergence to Limits via Approximations

• Verify limits by plugging in close values:
  • Example: $4.1, 4.01$ show convergence to $\frac{-1}{16}$

Limit Involving Square Roots

• Problem: Find $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$:
• Multiply by the conjugate $\sqrt{x} + 3$.
• Result after simplification yields final result: $\frac{1}{6}$.

Evaluating Limits Graphically

• One-sided limits:

  • Example of approach to $-3$:
  • 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.