Exploring Types of Discontinuities - AP Calculus AB

1. Let $g(x)=\dfrac{1}{x-2}$. What type of discontinuity does $g$ have at $x=2$?
   Answer: Infinite discontinuity The function g(x) = 1/(x-2) has an infinite discontinuity at x = 2. As x approaches 2 from the right, the denominator approaches 0 through positive values, making g(x) approach +∞. As x approaches 2 from the left, the denominator approaches 0 through negative values, making g(x) approach -∞. Since at least one of the one-sided limits is infinite, this is an infinite discontinuity (also called a vertical asymptote). Students sometimes confuse this with a removable discontinuity, but removable discontinuities require finite limits. To identify infinite discontinuities: look for denominators that approach zero without a corresponding zero in the numerator that would create a cancellable factor.
2. For $q(x)=\dfrac{|x|}{x}$, what type of discontinuity does $q$ have at $x=0$?
   Answer: Jump discontinuity The function $q(x)=\frac{|x|}{x}$ has a jump discontinuity at $x=0$. For $x<0$, we have $|x|=-x$, so $q(x)=\frac{-x}{x}=-1$. For $x>0$, we have $|x|=x$, so $q(x)=\frac{x}{x}=1$. Therefore, $\lim_{x\to 0^-}q(x)=-1$ and $\lim_{x\to 0^+}q(x)=1$. The function is undefined at $x=0$ due to division by zero. Since the one-sided limits exist but are different (-1 vs 1), this is a jump discontinuity. Students might think this is infinite because of division by zero, but the absolute value in the numerator ensures finite one-sided limits. A jump discontinuity is characterized by finite but unequal one-sided limits.
3. For $\nu(x)=\frac{x^2-2x-3}{x+1}$, what type of discontinuity does $\nu$ have at $x=-1$?
   Answer: Removable discontinuity This function has a removable discontinuity at x = -1. The expression ν(x) = (x²-2x-3)/(x+1) factors as ν(x) = (x-3)(x+1)/(x+1) = x-3 for x ≠ -1. The limit as x approaches -1 is lim(x→-1) (x-3) = -4, but the function is undefined at x = -1. This creates a removable discontinuity because we could define ν(-1) = -4 to make the function continuous. Students should factor the quadratic numerator to find (x-3)(x+1), allowing the (x+1) factors to cancel. To identify removable discontinuities: factor polynomials completely and look for common factors that create 0/0 forms with finite limits.
4. For $\eta(x)=\frac{x^2-4}{x-2}$, what type of discontinuity does $\eta$ have at $x=2$?
   Answer: Removable discontinuity This function has a removable discontinuity at x = 2. The expression η(x) = (x²-4)/(x-2) = (x-2)(x+2)/(x-2) simplifies to η(x) = x+2 for all x ≠ 2, but η(2) is undefined since we get 0/0. The limit as x approaches 2 is lim(x→2) (x+2) = 4, but the function is undefined at x = 2. This creates a removable discontinuity because we could define η(2) = 4 to make the function continuous. Students often recognize this as a standard difference of squares factorization. To identify removable discontinuities: look for factorable expressions where common factors in numerator and denominator cancel to yield finite limits.
5. Let $p(x)=\dfrac{x^2-4}{x-2}$. What type of discontinuity does $p$ have at $x=2$?
   Answer: Removable discontinuity The function p(x) = (x²-4)/(x-2) has a removable discontinuity at x = 2. We can factor the numerator as (x-2)(x+2), allowing us to cancel the (x-2) term for x ≠ 2, giving p(x) = x+2. The limit as x approaches 2 is therefore 2+2 = 4, which exists and is finite. However, p(2) is undefined because of division by zero in the original expression. Since the limit exists but the function is undefined at that point, this is a removable discontinuity—we could "fill the hole" by defining p(2) = 4. Students often mistake this for an infinite discontinuity because of the division by zero, but the key is that the zero in the denominator is cancelled by a corresponding factor in the numerator. To classify: if you can algebraically simplify to eliminate the problematic point, it's removable.
6. For $w(x)=\frac{x^2-4x}{x}$, what type of discontinuity does $w$ have at $x=0$?
   Answer: Removable discontinuity This function has a removable discontinuity at x = 0. The expression w(x) = (x²-4x)/x = x(x-4)/x simplifies to w(x) = x-4 for all x ≠ 0, but w(0) is undefined since we get 0/0. The limit as x approaches 0 is lim(x→0) (x-4) = -4, but the function is undefined at x = 0. This creates a removable discontinuity because we could define w(0) = -4 to make the function continuous. Students often think division by zero always creates infinite discontinuities, but when both numerator and denominator approach zero, check for simplification. To identify removable discontinuities: look for 0/0 forms that simplify to give finite limits.
7. For $f(x)=\begin{cases}\frac{x^2-9}{x-3},&x\ne3\\2,&x=3\end{cases}$, what type of discontinuity does $f$ have at $x=3$?
   Answer: Removable discontinuity This function has a removable discontinuity at x = 3. When x ≠ 3, we can factor the numerator as (x-3)(x+3), so f(x) = (x-3)(x+3)/(x-3) = x+3 for x ≠ 3. The limit as x approaches 3 is lim(x→3) (x+3) = 6, but f(3) = 2. Since the limit exists but doesn't equal the function value, this creates a removable discontinuity. Students often confuse this with an infinite discontinuity because of the rational form, but the key is that the factor (x-3) cancels from numerator and denominator. To classify discontinuities: check if the limit exists and compare it to the function value at that point.
8. Let $g(x)=\begin{cases}x+2,&x<1\\5-x,&x\ge 1\end{cases}$. What type of discontinuity, if any, occurs at $x=1$?
   Answer: Jump discontinuity For the piecewise function g(x), we need to check the left and right limits at x=1. From the left: lim(x→1⁻) g(x) = 1+2 = 3. From the right: lim(x→1⁺) g(x) = 5-1 = 4. Since the left limit (3) doesn't equal the right limit (4), the overall limit doesn't exist at x=1. When one-sided limits exist but are unequal, this creates a jump discontinuity. Students often confuse this with a removable discontinuity, but removable discontinuities require the two-sided limit to exist. The key strategy: for piecewise functions, always check both one-sided limits at transition points—if they're finite but different, it's a jump discontinuity.
9. Let $\delta(x)=\begin{cases}\frac{1}{x},&x<0\\1,&x\ge0\end{cases}$. What type of discontinuity does $\delta$ have at $x=0$?
   Answer: Infinite discontinuity This function has an infinite discontinuity at $x = 0$. The left-hand limit is $\lim_{x \to 0^{-}} \frac{1}{x} = -\infty$, while the right-hand limit is $\lim_{x \to 0^{+}} 1 = 1$, and $\delta(0) = 1$. Since the left-hand limit is infinite, this creates an infinite discontinuity. The function exhibits vertical asymptote behavior as x approaches 0 from the left. Students might focus on the fact that the right limit and function value exist, but an infinite discontinuity occurs when at least one one-sided limit is infinite. To identify infinite discontinuities: check if either one-sided limit approaches $\pm \infty$, regardless of the other limit's behavior.
10. For $v(x)=\dfrac{x^2-16}{x-4}$ when $x\ne 4$ and $v(4)=9$, what type of discontinuity occurs at $x=4$?
    Answer: Removable discontinuity The function v(x) = (x²-16)/(x-4) can be factored as (x+4)(x-4)/(x-4), which simplifies to x+4 for x≠4. Thus, lim(x→4) v(x) = 4+4 = 8. However, v(4) is defined as 9, which differs from the limit value of 8. Since the limit exists (is finite) but doesn't equal the function value, this is a removable discontinuity at x=4. The discontinuity could be "removed" by redefining v(4) = 8 to make the function continuous. Students often overlook that even when a function is defined at a point, if that value doesn't match the limit, a removable discontinuity exists. Classification tip: removable discontinuities occur when limits exist but don't match function values.
11. Let $g(x)=\begin{cases}x+1,&x<0\\3,&x\ge0\end{cases}$. What type of discontinuity does $g$ have at $x=0$?
    Answer: Jump discontinuity This function has a jump discontinuity at x = 0. The left-hand limit is lim(x→0⁻) (x+1) = 1, while the right-hand limit is lim(x→0⁺) 3 = 3, and g(0) = 3. Since the left and right limits exist but are unequal (1 ≠ 3), this creates a jump discontinuity. The function literally 'jumps' from approaching 1 on the left to the value 3 on the right. Students might think this is continuous because g(0) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: check that both one-sided limits exist but are different values.
12. Let $Z(x)=\begin{cases}3x,&x<2\\6,&x\ge2\end{cases}$. What type of discontinuity does $Z$ have at $x=2$?
    Answer: No discontinuity (continuous) This function has no discontinuity at x = 2. The left-hand limit is lim(x→2⁻) 3x = 6, the right-hand limit is lim(x→2⁺) 6 = 6, and Z(2) = 6. Since both one-sided limits exist, are equal, and match the function value, the function is continuous at x = 2. This is an example of a piecewise function that is carefully constructed to be continuous at the boundary point. Students might assume piecewise functions are automatically discontinuous, but continuity depends on matching limits and function values. To verify continuity: check that both one-sided limits exist, are equal, and match the function value.
13. For $h(x)=\begin{cases}x+2,&x<1\\x^2,&x\ge 1\end{cases}$, what type of discontinuity does $h$ have at $x=1$?
    Answer: Jump discontinuity The function $h(x)$ has a jump discontinuity at $x=1$. From the left, $\lim_{x\to 1^-}h(x)=\lim_{x\to 1^-}(x+2)=3$. From the right, $\lim_{x\to 1^+}h(x)=\lim_{x\to 1^+}(x^2)=1$. At $x=1$, we use the second formula since $1\geq 1$, so $h(1)=1^2=1$. The left and right limits exist but are different (3 vs 1), which is the defining characteristic of a jump discontinuity. Students might think this is removable since $h(1)$ equals one of the limits, but removable discontinuities require both one-sided limits to be equal. A jump discontinuity occurs when both one-sided limits exist as finite values but are not equal to each other.
14. For $P(x)=\frac{1}{(x-1)(x+2)}$, what type of discontinuity does $P$ have at $x=-2$?
    Answer: Infinite discontinuity This function has an infinite discontinuity at x = -2. As x approaches -2, the denominator (x-1)(x+2) approaches 0 while the numerator remains 1, causing the function values to approach ±∞. The factor (x+2) in the denominator creates a vertical asymptote at x = -2. This defines an infinite discontinuity because the function approaches infinite values as x approaches -2. Students might confuse this with the behavior at x = 1, but each zero of the denominator creates its own infinite discontinuity. To identify infinite discontinuities: check each value that makes the denominator zero while keeping the numerator non-zero.
15. Let $t(x)=\begin{cases}\frac{\sin x}{x},&x\ne0\\1,&x=0\end{cases}$. What type of discontinuity occurs at $x=0$?
    Answer: No discontinuity (continuous) This function has no discontinuity at x = 0. The limit of sin(x)/x as x approaches 0 is a fundamental limit equal to 1, and t(0) is defined as 1. Since lim(x→0) sin(x)/x = 1 = t(0), the function is continuous at x = 0. This is a classic example where a rational expression that's undefined at a point can be made continuous by defining the function value to equal the limit. Students often think this is a removable discontinuity because sin(x)/x is undefined at x = 0, but since t(0) = 1 matches the limit, there's no discontinuity. To determine continuity: check that the limit exists and equals the function value at the point.
16. Let $g(x)=\begin{cases}x^2,&x\ne2\\7,&x=2\end{cases}$. What type of discontinuity does $g$ have at $x=2$?
    Answer: Removable discontinuity This function has a removable discontinuity at x = 2. Since g(x) = x² for x ≠ 2, the limit as x approaches 2 is lim(x→2) x² = 4, but g(2) = 7. Since the limit exists but doesn't equal the function value (4 ≠ 7), this creates a removable discontinuity. The discontinuity could be 'removed' by redefining g(2) = 4. Students might think this is continuous because x² is a polynomial, but the piecewise definition changes the value at x = 2. To identify removable discontinuities: check if the limit exists but differs from the function value at the point.
17. Let $\beta(x)=\frac{x^2-9}{x^2-3x}$. What type of discontinuity does $\beta$ have at $x=0$?
    Answer: Infinite discontinuity This function has an infinite discontinuity at x = 0. The expression β(x) = (x²-9)/(x²-3x) = (x²-9)/(x(x-3)) has a denominator that equals zero when x = 0, while the numerator equals -9 ≠ 0. As x approaches 0, the function approaches ±∞, creating a vertical asymptote. This defines an infinite discontinuity. Students might confuse this with the behavior at x = 3, but each zero of the denominator must be analyzed separately. To identify infinite discontinuities: check points where denominators approach zero while numerators approach non-zero values.
18. Let $c(x)=\begin{cases}0,&x<3\\1,&x\ge3\end{cases}$. What type of discontinuity does $c$ have at $x=3$?
    Answer: Jump discontinuity This function has a jump discontinuity at x = 3. The left-hand limit is lim(x→3⁻) 0 = 0, while the right-hand limit is lim(x→3⁺) 1 = 1, and c(3) = 1. Since both one-sided limits exist but are unequal (0 ≠ 1), this creates a jump discontinuity. The function jumps from 0 on the left to 1 on the right at x = 3. Students might think this is continuous because c(3) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.
19. For $\epsilon(x)=\frac{(x-2)^2}{x-2}$, what type of discontinuity does $\epsilon$ have at $x=2$?
    Answer: Removable discontinuity This function has a removable discontinuity at x = 2. The expression ε(x) = (x-2)²/(x-2) simplifies to ε(x) = x-2 for all x ≠ 2, but ε(2) is undefined since we get 0/0. The limit as x approaches 2 is lim(x→2) (x-2) = 0, but the function is undefined at x = 2. This creates a removable discontinuity because we could define ε(2) = 0 to make the function continuous. Students might not recognize that (x-2)² = (x-2)(x-2), allowing one factor to cancel. To identify removable discontinuities: factor expressions completely, including repeated factors, and look for cancellation opportunities.
20. For $S(x)=\begin{cases}x-2,&x<2\\0,&x=2\\x+1,&x>2\end{cases}$, what type of discontinuity occurs at $x=2$?
    Answer: Jump discontinuity This function has a jump discontinuity at x = 2. The left-hand limit is lim(x→2⁻) (x-2) = 0, the function value is S(2) = 0, and the right-hand limit is lim(x→2⁺) (x+1) = 3. Since the left and right limits exist but are unequal (0 ≠ 3), this creates a jump discontinuity. The function jumps from 0 on the left to 3 on the right at x = 2. Students might think this is continuous because S(2) equals the left-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.

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