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.

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.

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.

This function has a removable discontinuity at x = 3. The expression T(x) = (x-3)(x+1)/(x-3) simplifies to T(x) = x+1 for all x ≠ 3, but T(3) is undefined since we get 0/0. The limit as x approaches 3 is lim(x→3) (x+1) = 4, but the function is undefined at x = 3. This creates a removable discontinuity because we could define T(3) = 4 to make the function continuous. Students often see the rational form and assume infinite discontinuity, but factor cancellation is key. To identify removable discontinuities: look for common factors in numerator and denominator that create 0/0 forms.

This function has a removable discontinuity at x = -4. The expression r(x) = (x+4)/(x+4) simplifies to r(x) = 1 for all x ≠ -4, but r(-4) is undefined since we get 0/0. The limit as x approaches -4 is lim(x→-4) 1 = 1, but the function is undefined at x = -4. This creates a removable discontinuity because we could define r(-4) = 1 to make the function continuous. Students often think this is an infinite discontinuity because of the rational form, but the key is that both numerator and denominator approach zero simultaneously. To identify removable discontinuities: look for 0/0 indeterminate forms that can be simplified to find a finite limit.

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.

This function has a removable discontinuity at x = 3. The expression V(x) = (x²-6x+9)/(x-3) = (x-3)²/(x-3) simplifies to V(x) = x-3 for all x ≠ 3, but V(3) is undefined since we get 0/0. The limit as x approaches 3 is lim(x→3) (x-3) = 0, but the function is undefined at x = 3. This creates a removable discontinuity because we could define V(3) = 0 to make the function continuous. Students might not recognize the perfect square factorization, but the key is that both numerator and denominator have the factor (x-3). To identify removable discontinuities: factor completely and look for common factors that cancel.

This function has a removable discontinuity at x = 5. The expression d(x) = (x-5)/((x-5)(x+1)) simplifies to d(x) = 1/(x+1) for all x ≠ 5, but d(5) is undefined since we get 0/0. The limit as x approaches 5 is lim(x→5) 1/(x+1) = 1/6, but the function is undefined at x = 5. This creates a removable discontinuity because we could define d(5) = 1/6 to make the function continuous. Students often miss the factor cancellation, thinking this is an infinite discontinuity. To identify removable discontinuities: look for common factors in numerator and denominator that create 0/0 forms.

This function has an infinite discontinuity at x = 1. The expression ι(x) = 1/(x²-1) = 1/((x-1)(x+1)) has a denominator that approaches 0 as x approaches 1, while the numerator remains 1. This causes the function to approach ±∞, creating a vertical asymptote at x = 1. The sign of the denominator determines whether the function approaches +∞ or -∞ from each side. This defines an infinite discontinuity. Students might confuse this with the behavior at x = -1, but both zeros of the denominator create infinite discontinuities. To identify infinite discontinuities: find values where denominators approach zero while numerators approach non-zero values.

The function p(x) = (x²-4x)/x = x(x-4)/x simplifies to x-4 for x≠0. Therefore, lim(x→0) p(x) = 0-4 = -4. Since p(0) is defined as 7, which differs from the limit value of -4, we have a removable discontinuity at x=0. The discontinuity is called "removable" because we could "remove" it by redefining p(0) = -4 to make the function continuous. Students often assume any function undefined at a point must have an infinite discontinuity, but when simplification yields a finite limit, it's removable. To identify removable discontinuities: simplify the expression, find the limit, and check if it's finite but different from the function value.