Solved Examples: Proving Continuity at a Point in Pre-Calculus.

📚 Quick Study Guide

• 🔍 Definition: A function $f(x)$ is continuous at a point $x = a$ if the following three conditions are met:
  1. $f(a)$ is defined.
  2. $\lim_{x \to a} f(x)$ exists.
  3. $\lim_{x \to a} f(x) = f(a)$.
• ➕ Sum/Difference: If $f(x)$ and $g(x)$ are continuous at $x=a$, then $f(x) + g(x)$ and $f(x) - g(x)$ are also continuous at $x=a$.
• ✖️ Product: If $f(x)$ and $g(x)$ are continuous at $x=a$, then $f(x) \cdot g(x)$ is also continuous at $x=a$.
• ➗ Quotient: If $f(x)$ and $g(x)$ are continuous at $x=a$, and $g(a) \neq 0$, then $\frac{f(x)}{g(x)}$ is continuous at $x=a$.
• 💡 Polynomials: Polynomial functions are continuous everywhere.
• 📈 Rational Functions: Rational functions are continuous everywhere except where the denominator is zero.
• 🧭 Piecewise Functions: Pay special attention to the points where the function definition changes; check continuity at these points.

🧪 Practice Quiz

1. Question 1: Determine if the function $f(x) = \begin{cases} x^2, & x < 1 \\ 2x - 1, & x \geq 1 \end{cases}$ is continuous at $x = 1$.
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(1)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 1} f(x)$ does not exist.
4. D) No, it is not continuous because $\lim_{x \to 1} f(x) \neq f(1)$.


2. Question 2: Is the function $f(x) = \frac{x+2}{x-3}$ continuous at $x = 3$?
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(3)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 3} f(x)$ exists.
4. D) It cannot be determined.


3. Question 3: Determine if $f(x) = x^3 - 2x + 1$ is continuous at $x = 2$.
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(2)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 2} f(x)$ does not exist.
4. D) No, it is not continuous because $\lim_{x \to 2} f(x) \neq f(2)$.


4. Question 4: Examine the continuity of the function $f(x) = \begin{cases} 3x + 1, & x < 0 \\ x - 2, & x \geq 0 \end{cases}$ at $x = 0$.
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(0)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 0} f(x)$ does not exist.
4. D) No, it is not continuous because $\lim_{x \to 0} f(x) = f(0)$.


5. Question 5: Is the function $f(x) = |x|$ continuous at $x = 0$?
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(0)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 0} f(x)$ does not exist.
4. D) No, it is not continuous because $\lim_{x \to 0} f(x) \neq f(0)$.


6. Question 6: Is the function $f(x) = \frac{x^2 - 4}{x - 2}$ continuous at $x = 2$?
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(2)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 2} f(x)$ does not exist.
4. D) No, it is not continuous because $\lim_{x \to 2} f(x) = f(2)$.


7. Question 7: Determine if $f(x) = \begin{cases} x + 1, & x < 2 \\ 5 - x, & x \geq 2 \end{cases}$ is continuous at $x = 2$.
1. A) Yes, it is continuous.
2. B) No, it is not continuous because $f(2)$ is not defined.
3. C) No, it is not continuous because $\lim_{x \to 2} f(x)$ does not exist.
4. D) No, it is not continuous because $\lim_{x \to 2} f(x) \neq f(2)$.