t(x) shows removable discontinuity at x=8 via (x-8). Set to limit: simplifies to x+8, 8+8=16. Replace 0 with 16, choice B. Confusion from direct sub: undefined. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

f(x) shows removable discontinuity at x=5. Simplifies: (x-5)(x+3)/(x-5) = x+3, limit 5+3=8. Replace 12 with 8, choice C. Confusion: 0/0. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

u(x) has removable discontinuity at x=9. Limit: x+9, 9+9=18. Replace 100 with 18, choice A. Common error: 0/0 without simplifying. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

The function has a removable discontinuity at x = 3 because the numerator x² - 9 = (x - 3)(x + 3) contains the factor (x - 3) that cancels with the denominator. After canceling, we get f(x) = x + 3 for x ≠ 3. To find the value that removes the discontinuity, we calculate lim(x→3) f(x) = lim(x→3) (x + 3) = 3 + 3 = 6. A common mistake is to think the function equals 0 at the discontinuity or to use the current defined value of 5. The strategy is: factor the numerator, cancel common factors with the denominator, then evaluate the simplified expression 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.

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.

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 v(x) has a removable discontinuity at x = 7 because the numerator x² - 49 = (x - 7)(x + 7) contains the factor (x - 7) that cancels with the denominator. After canceling, we get v(x) = x + 7 for x ≠ 7. To remove the discontinuity, we need v(7) = lim(x→7) (x + 7) = 7 + 7 = 14. Students might keep the current value of 12 or forget to add when evaluating. The key strategy is recognizing difference of squares (a² - b² = (a-b)(a+b)), canceling common factors, then evaluating the simplified expression.

The function p(x) = (x² - 16)/(x - 4) has a removable discontinuity at x = 4 because x² - 16 factors as (x - 4)(x + 4). After canceling (x - 4), we get p(x) = x + 4 for x ≠ 4. To remove the discontinuity, p(4) should equal the limit: lim[x→4] (x + 4) = 4 + 4 = 8. The current value p(4) = 0 creates a jump discontinuity, but setting p(4) = 8 makes the function continuous. Students sometimes think that because the function is undefined at x = 4 in its original form, no value can make it continuous, but factoring reveals the removable nature. The strategy is to recognize difference of squares patterns, factor completely, cancel, and evaluate the simplified expression.

The function r(x) = (x² - 25)/(x + 5) has a removable discontinuity at x = -5 because x² - 25 factors as (x - 5)(x + 5). After canceling (x + 5), we get r(x) = x - 5 for x ≠ -5. To remove the discontinuity, r(-5) should equal the limit: lim[x→-5] (x - 5) = -5 - 5 = -10. The current value r(-5) = 1 creates a jump discontinuity, but setting r(-5) = -10 makes the function continuous. Students often make sign errors when substituting negative values or forget that x + 5 appears in the factorization of x² - 25. The key strategy is to recognize and factor difference of squares, cancel the common factor, then carefully evaluate at the point of discontinuity.