Removing Discontinuities - AP Calculus BC

1. What is the behavior of $f(x)$ at a jump discontinuity?

   Differing left and right-hand limits. Left and right limits exist but are unequal.

2. What kind of discontinuity does $f(x) = \frac{1}{x}$ have at $x = 0$?

   Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.

3. Determine the discontinuity type: $f(x) = \frac{1}{x}$ at $x = 0$.

   Infinite discontinuity. Division by zero with no cancellation creates a vertical asymptote.

4. Identify the discontinuity type: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.

   Removable discontinuity. The factor $(x-2)$ cancels, creating a hole at $x=2$.

5. What is the general approach to remove a discontinuity?

   Simplify and redefine the function. Factor out common terms and define the function at the hole.

6. Identify the discontinuity at $x = 0$ for $f(x) = \frac{1}{x^2}$.

   Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.

7. What is the limit $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$?

   3. Factor: $\frac{(x^2+x+1)(x-1)}{x-1} = x^2+x+1$, so limit is 3.

8. Find the removable discontinuity: $f(x) = \frac{x^2 - 9}{x - 3}$.

   $x = 3$. The denominator equals zero when $x=3$, creating a hole.

9. What condition creates a removable discontinuity?

   A factor cancels in the numerator and denominator. Common factors in numerator and denominator create holes.

10. What does it mean for a function to be continuous?

    The function is continuous if it's defined and its limit equals the function value. Three conditions: defined, limit exists, and they're equal.

11. What is a removable discontinuity in a function?

    A point where a function is not defined but can be redefined to make it continuous. The limit exists but doesn't equal the function value.

12. What is the limit as $x \to 3$ of $\frac{x^2 - 9}{x - 3}$?

    6. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.

13. Find and remove the discontinuity: $f(x) = \frac{x^2 - 4x}{x - 4}$.

    Redefine $f(4) = 4$. Factor: $\frac{x(x-4)}{x-4} = x$, so limit at $x=4$ is 4.

14. How do you remove a removable discontinuity?

    Redefine the function at the discontinuity point. Set the function value equal to the limit at that point.

15. State the condition for a function to have no discontinuities.

    The function must be continuous everywhere in its domain. No holes, jumps, or vertical asymptotes anywhere.

16. What is the condition for a removable discontinuity?

    A cancelable factor in the expression. A common factor exists in both numerator and denominator.

17. What is an infinite discontinuity?

    A discontinuity where a function approaches infinity at a point. The function has a vertical asymptote at that point.

18. What is the limit of $\lim_{x \to 1} \frac{x^2 - 1}{x - 1}$?

    2. Factor: $\frac{(x+1)(x-1)}{x-1} = x+1$, so limit is $1+1=2$.

19. Which discontinuity is present in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x=1$?

    Removable. The factor $(x-1)$ cancels, leaving a hole at $x=1$.

20. What happens to $f(x)$ at an essential discontinuity?

    The limits don't exist or are undefined. The function oscillates wildly or has no pattern near the point.

21. What is the limit as $x \to 2$ of $\frac{x^2 - 4}{x - 2}$?

    4. Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$.

22. Find the limit: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.

    4. Factor and cancel: $\frac{(x+2)(x-2)}{x-2} = x+2$, so limit is $2+2=4$.

23. When is a discontinuity considered non-removable?

    If limits do not exist or differ at a point. No common factors cancel or limits become infinite.

24. What must be true for $f(x)$ to be continuous at $x = a$?

    $\lim_{x \to a} f(x) = f(a)$. All three continuity conditions must hold at the point.

25. Identify the removable discontinuity in $f(x) = \frac{x^2 - 4x}{x(x - 4)}$.

    $x = 0$. The factor $x$ in denominator creates a hole when $x=0$.

26. What does $\lim_{x \to a} f(x) = \infty$ imply about $f(x)$ at $x = a$?

    Infinite discontinuity at $x = a$. The function approaches infinity, creating a vertical asymptote.

27. What is the limit definition of continuity at $x = a$?

    $\lim_{x \to a} f(x) = f(a)$. The limit as $x$ approaches $a$ equals the function value.

28. What is a jump discontinuity?

    A discontinuity where a function has differing left and right limits at a point. The function 'jumps' from one value to another at the point.

29. Remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.

    Redefine $f(2) = 4$. The limit at $x=2$ is 4, so define $f(2)=4$ for continuity.

30. Identify and remove the discontinuity of $f(x) = \frac{x^2 - 9}{x - 3}$.

    Redefine $f(3) = 6$. Factor: $\frac{(x+3)(x-3)}{x-3} = x+3$, so limit is 6.

31. Identify the discontinuity type: $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$.

    Removable discontinuity. The factor $(x-1)$ cancels, creating a hole at $x=1$.

32. What type of discontinuity occurs at $f(x) = \frac{1}{x}$ at $x = 0$?

    Infinite discontinuity. Division by zero creates a vertical asymptote at $x=0$.

33. How is a jump discontinuity characterized?

    Differing left and right-hand limits. The function has different values approaching from each side.

34. Determine the discontinuity: $f(x) = \frac{x^2 - 4}{x - 2}$ at $x = 2$.

    Removable discontinuity. The factor $ (x-2) $ cancels, leaving a hole at $x=2$.

35. What is the necessary condition for a function to be continuous?

    The function is defined and limits match the function value. Function exists, limit exists, and both values are equal.

36. Find the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$.

    $x = 5$. The denominator becomes zero when $x=5$, creating a hole.

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