p(x) has removable discontinuity at x=2. Factoring: (x-2)(x-3)/(x-2) = x-3, limit 2-3=-1. Replace 1 with -1, choice A. Mistake: direct plug. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

p(x) shows a removable discontinuity at x=4 as (x²-16) and (x-4) share (x-4). Set p(4) to the limit for continuity. Simplifying to (x-4)(x+4)/(x-4) = x+4. Limit at 4 is 4+4=8. Replace 10 with 8, choice A. Confusion arises from unsimplified substitution, giving undefined, but factoring resolves it. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

s(x) has a removable discontinuity at x=7 with shared (x-7). Continuity needs limit value. Simplifies to x+7, limit 7+7=14. Replace 20 with 14, choice A. Mistake: plugging without simplifying gives 0/0. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

For q(x), the discontinuity at x=6 is removable because x² - 2x - 24 factors to (x-6)(x+4), canceling to x+4. The limit at x=6 is 6+4 = 10, the value to use for continuity. This fills the gap seamlessly. A common confusion is choosing -4 or another coefficient incorrectly. Some overlook the positive sign in the factor. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.

q(x) has a removable discontinuity at x=5 due to the shared factor (x-5) in numerator and denominator. For continuity, q(5) should be the limit. Simplifies to x+5, limit 5+5=10. Replace 0 with 10, choice C. Direct plug-in confuses by giving 0/0, but simplification shows 10. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

The function r(x) has a removable discontinuity at x = 5 because the numerator x² - 25 = (x - 5)(x + 5) contains the factor (x - 5) that cancels with the denominator. After canceling, we get r(x) = x + 5 for x ≠ 5. To remove the discontinuity, we need r(5) = lim(x→5) (x + 5) = 5 + 5 = 10. Students often confuse this with the current value of 11 or try to evaluate the original fraction directly. The key insight is that removable discontinuities can be fixed by finding the limit of the simplified function after canceling common factors.

For s(x), there is a removable discontinuity at x=3 as the numerator x² - 10x + 21 factors into (x-3)(x-7), allowing cancellation with x-3. This simplifies to x-7 for x ≠ 3. To make s continuous, set s(3) to the limit of x-7 as x approaches 3, yielding 3-7 = -4. This fills the gap, ensuring no break in the graph. A common confusion arises from directly substituting x=3 into the unsimplified form, which is undefined, instead of simplifying. Some mix up the signs when evaluating the limit. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.

The function q(x) exhibits a removable discontinuity at x=1 since the numerator x² - 6x + 5 factors into (x-1)(x-5), sharing a common factor with the denominator x-1. Simplifying yields x-5 for x ≠ 1. To remove the discontinuity, define q(1) as the limit of x-5 as x approaches 1, which is -4. This makes the function continuous at x=1 by filling the hole in the graph. A common confusion is attempting to plug x=1 directly into the original expression, resulting in an undefined value, rather than simplifying first. Another mistake is confusing the value with the roots of the numerator. As a transferable strategy, always factor the numerator and denominator of rational functions, cancel common factors, and set the function value at the discontinuity to the limit of the simplified expression.

The function r(x) has a removable discontinuity at x = -5 where (x² - 25)/(x + 5) is undefined. Factoring the numerator as a difference of squares gives x² - 25 = (x + 5)(x - 5), so the expression simplifies to (x + 5)(x - 5)/(x + 5) = x - 5 for x ≠ -5. The limit as x approaches -5 is lim[x→-5] (x - 5) = -5 - 5 = -10. To remove the discontinuity, we must replace 9 with -10. Students often make sign errors when substituting negative values. Always factor completely, cancel common factors, then carefully evaluate the limit at the point of discontinuity.

The function f(x) = (x² - 9)/(x - 3) has a removable discontinuity at x = 3 because the numerator factors as (x - 3)(x + 3), allowing us to cancel the (x - 3) term. After cancellation, we get f(x) = x + 3 for x ≠ 3. To find the value that removes the discontinuity, we calculate the limit as x approaches 3: lim[x→3] (x + 3) = 3 + 3 = 6. The current value f(3) = 8 creates a jump discontinuity, but defining f(3) = 6 would make the function continuous. A common mistake is thinking the discontinuity cannot be removed because division by zero seems problematic, but factoring reveals the cancellation possibility. The strategy is: factor the numerator, cancel common factors with the denominator, then evaluate the simplified expression at the point of discontinuity.