Limits, Piecewise Functions, and Continuity Study Guide

Analysis of Piecewise Function 1

• Piecewise Definition:

  • $f(x) = \begin{cases} 5x + 3 & \text{when } x < 2 \\ 2x^2 + 5 & \text{when } 2 \le x < 4 \\ x^3 - 5x + 3 & \text{when } x \ge 4 \end{cases}$

• One-Sided and Two-Sided Limits at Boundary $x = 2$:

  • Left-Hand Limit Calculation ($\lim_{x \to 2^-} f(x)$):
  • When approaching $x = 2$ from the left, select values less than $2$ (for instance, $x = 1.99$ rather than $x = 2.01$ on a number line).
  • Use the piecewise expression corresponding to $x < 2$, which is $5x + 3$
  • Evaluate by direct substitution:
    • $\lim_{x \to 2^-} f(x) = 5(2) + 3 = 10 + 3 = 13$
  • Right-Hand Limit Calculation ($\lim_{x \to 2^+} f(x)$):
  • When approaching $x = 2$ from the right, select values greater than $2$.
  • Use the piecewise expression corresponding to $2 \le x < 4$, which is $2x^2 + 5$
  • Evaluate by direct substitution:
    • $\lim_{x \to 2^+} f(x) = 2(2)^2 + 5 = 2(4) + 5 = 8 + 5 = 13$
  • Two-Sided Limit Determination ($\lim_{x \to 2} f(x)$):
  • Since the left-hand limit and right-hand limit are equal ($13 = 13$), the two-sided limit exists.
  • $\lim_{x \to 2} f(x) = 13$

• Function Value at $x = 2$ ($f(2)$):

  • Determine which piecewise inequality includes $x = 2$
  • The inequality $2 \le x < 4$ contains $x = 2$, requiring the use of $2x^2 + 5$
  • Evaluate:
  • $f(2) = 2(2)^2 + 5 = 8 + 5 = 13$
  • The function is defined at $x = 2$ with $f(2) = 13$

• One-Sided and Two-Sided Limits at Boundary $x = 4$:

  • Left-Hand Limit Calculation ($\lim_{x \to 4^-} f(x)$):
  • For values approaching $4$ from the left ($x < 4$), use the function piece $2x^2 + 5$
  • Evaluate:
    • $\lim_{x \to 4^-} f(x) = 2(4)^2 + 5 = 2(16) + 5 = 32 + 5 = 37$
  • Right-Hand Limit Calculation ($\lim_{x \to 4^+} f(x)$):
  • For values approaching $4$ from the right ($x > 4$), use the function piece $x^3 - 5x + 3$
  • Evaluate:
    • $\lim_{x \to 4^+} f(x) = 4^3 - 5(4) + 3 = 64 - 20 + 3 = 47$
  • Two-Sided Limit Determination ($\lim_{x \to 4} f(x)$):
  • The left-hand limit ($37$) and right-hand limit ($47$) do not match ($37 \neq 47$).
  • Consequently, the two-sided limit does not exist:
    • $\lim_{x \to 4} f(x) = \text{Does Not Exist (DNE)}$

• Function Value at $x = 4$ ($f(4)$):

  • The inequality $x \ge 4$ explicitly includes $x = 4$, designating $x^3 - 5x + 3$
  • Evaluate:
  • $f(4) = 4^3 - 5(4) + 3 = 64 - 20 + 3 = 47$

Analysis of Piecewise Function 2

• Piecewise Definition:

  • $f(x) = \begin{cases} 7x - 5 & \text{when } x < 1 \\ 5 & \text{when } x = 1 \\ 3x^2 - x & \text{when } 1 < x \le 3 \\ x^3 + 4 & \text{when } x > 3 \end{cases}$

• Evaluation of Limits and Function Value at Boundary $x = 1$:

  • Left-Hand Limit Calculation ($\lim_{x \to 1^-} f(x)$):
  • For $x < 1$, apply $7x - 5$
  • Evaluate:
    • $\lim_{x \to 1^-} f(x) = 7(1) - 5 = 7 - 5 = 2$
  • Right-Hand Limit Calculation ($\lim_{x \to 1^+} f(x)$):
  • For $1 < x \le 3$, apply $3x^2 - x$
  • Evaluate:
    • $\lim_{x \to 1^+} f(x) = 3(1)^2 - 1 = 3(1) - 1 = 2$
  • Two-Sided Limit Determination ($\lim_{x \to 1} f(x)$):
  • Since both one-sided limits equal $2$, the two-sided limit exists:
    • $\lim_{x \to 1} f(x) = 2$
  • Exact Function Value Calculation ($f(1)$):
  • The piecewise definition specifies $f(x) = 5$ at $x = 1$
  • Thus, $f(1) = 5$

• Evaluation of Limits and Function Value at Boundary $x = 3$:

  • Left-Hand Limit Calculation ($\lim_{x \to 3^-} f(x)$):
  • For $x < 3$ (specifically $1 < x \le 3$), apply $3x^2 - x$
  • Evaluate:
    • $\lim_{x \to 3^-} f(x) = 3(3)^2 - 3 = 3(9) - 3 = 27 - 3 = 24$
  • Right-Hand Limit Calculation ($\lim_{x \to 3^+} f(x)$):
  • For $x > 3$, apply $x^3 + 4$
  • Evaluate:
    • $\lim_{x \to 3^+} f(x) = 3^3 + 4 = 27 + 4 = 31$
  • Two-Sided Limit Determination ($\lim_{x \to 3} f(x)$):
  • Because the left-hand limit ($24$) and right-hand limit ($31$) are unequal, the limit does not exist:
    • $\lim_{x \to 3} f(x) = \text{Does Not Exist (DNE)}$
  • Exact Function Value Calculation ($f(3)$):
  • The inequality $1 < x \le 3$ includes $x = 3$, so apply $3x^2 - x$
  • Evaluate:
    • $f(3) = 3(3)^2 - 3 = 27 - 3 = 24$

Solving for Unknown Constants to Guarantee Continuity

• Problem Statement:

  • Given the piecewise function $f(x) = \begin{cases} Cx + 3 & \text{when } x < 2 \\ 3x + C & \text{when } x \ge 2 \end{cases}$
  • Find the value of the constant $C$ such that $f(x)$ is continuous at $x = 2$

• Theoretical Requirements for Continuity:

  • A function $f(x)$ is continuous at a point $x = a$ if and only if:
  1. $f(a)$ is defined.
  2. $\lim_{x \to a} f(x)$ exists (meaning $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$).
  3. $\lim_{x \to a} f(x) = f(a)$.
  • For this piecewise function to be continuous at $x = 2$, the expressions on either side of $x = 2$ must evaluate to the same value when $x = 2$

• Step-by-Step Algebraic Solution:

  • Step 1: Equate the two piecewise components:
  • $Cx + 3 = 3x + C$
  • Step 2: Substitute $x = 2$ into the equation:
  • $C(2) + 3 = 3(2) + C$
  • $2C + 3 = 6 + C$
  • Step 3: Isolate $C$ by subtracting $C$ from both sides:
  • $2C - C + 3 = 6$
  • $C + 3 = 6$
  • Step 4: Subtract $3$ from both sides to find $C$:
  • $C = 6 - 3$
  • $C = 3$

• Verification:

  • Substituting $C = 3$ into the left piece yields $3(2) + 3 = 9$
  • Substituting $C = 3$ into the right piece yields $3(2) + 3 = 9$
  • Both pieces yield $9$ at $x = 2$, confirming that $f(x)$ is continuous at $x = 2$ when $C = 3$