This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at $x=1$, we need to set $w(1)$ equal to the limit as $x$ approaches 1 of $w(x)$. By factoring the numerator as $(x-1)(x^2 + x + 1)$, we can cancel the $(x-1)$ term with the denominator, simplifying to $x^2 + x + 1$ for $x \neq 1$. Therefore, the limit as $x$ approaches 1 is $1+1+1=3$, which is the value that removes the discontinuity. A tempting distractor like 2 might be the originally given value of $w(1)$, but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.

This problem focuses on removing a removable discontinuity in a rational function. The function h(x) = (x² - 4x)/x has a discontinuity at x=0, but simplifying by factoring x(x-4)/x yields h(x) = x-4 for x ≠ 0. The limit as x approaches 0 is 0 - 4 = -4. Defining h(0) = -4 removes the discontinuity. A tempting distractor is 4, perhaps from ignoring the sign or evaluating |x-4|, but the correct simplified expression gives -4. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.

This problem focuses on removing a removable discontinuity in a rational function. The function f(x) = ((x-2)(x+2))/(x-2) simplifies to x + 2 for x ≠ 2 after canceling (x-2). The limit as x approaches 2 is 2 + 2 = 4. Setting f(2) = 4 removes the discontinuity. A tempting distractor is -1, the original value, but it ignores the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.

This problem focuses on removing a removable discontinuity in a rational function. The function f(x) = (x² - 9)/(x - 3) has a discontinuity at x=3 because the denominator is zero, but we can simplify by factoring the numerator as (x-3)(x+3) and canceling the common (x-3) factor, resulting in f(x) = x+3 for x ≠ 3. The limit as x approaches 3 is then 3 + 3 = 6. To remove the discontinuity, define f(3) = 6 to match this limit. A tempting distractor is 9, which comes from evaluating the numerator at x=3 without simplification, but it ignores the canceled factor creating the hole. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.

To remove the removable discontinuity at x = 4, we need q(4) to equal the limit. The function q(x) = (x²-16)/(x-4) can be simplified by factoring: x²-16 = (x-4)(x+4), so q(x) = (x-4)(x+4)/(x-4) = x+4 for x≠4. Therefore, lim[x→4] q(x) = 4+4 = 8, meaning q(4) should be 8 to make the function continuous. The given value q(4) = 0 doesn't match the limit and thus doesn't remove the discontinuity. Remember that difference of squares a²-b² = (a-b)(a+b) is a key factoring pattern for removing discontinuities.

This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=0, we need to set n(0) equal to the limit as x approaches 0 of n(x). By factoring the numerator as x(x-5), we can cancel the x term with the denominator, simplifying to x-5 for x ≠ 0. Therefore, the limit as x approaches 0 is 0-5=-5, which is the value that removes the discontinuity. A tempting distractor like 5 might come from ignoring the sign in the simplified expression. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.

This problem focuses on removing a removable discontinuity in a rational function. The function w(x) = (x² + x)/x simplifies to x + 1 for x ≠ 0 after canceling x. The limit as x approaches 0 is 0 + 1 = 1. Setting w(0) = 1 removes the discontinuity. A tempting distractor is 2, the original value, but it doesn't equal the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.

This problem focuses on removing a removable discontinuity in a rational function. The function $v(x) = \frac{x^2 - 36}{x - 6}$ factors to $\frac{(x-6)(x+6)}{x-6}$, simplifying to $x + 6$ for $x \neq 6$. The limit as $x$ approaches 6 is $6 + 6 = 12$. Setting $v(6) = 12$ removes the discontinuity. A tempting distractor is 1, the original value, but it ignores the limit. A transferable strategy is to simplify rational functions by canceling common factors before evaluating the limit to remove removable discontinuities.

This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=1, we need to set u(1) equal to the limit as x approaches 1 of u(x). By factoring the numerator as (x-1)(x+1), we can cancel the (x-1) term with the denominator, simplifying to x+1 for x ≠ 1. Therefore, the limit as x approaches 1 is 1+1=2, which is the value that removes the discontinuity. A tempting distractor like 0 might be the originally given value of u(1), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.

This problem involves identifying and removing a removable discontinuity in a rational function. To make the function continuous at x=10, we need to set v(10) equal to the limit as x approaches 10 of v(x). By factoring the numerator as (x-10)(x+10), we can cancel the (x-10) term with the denominator, simplifying to x+10 for x ≠ 10. Therefore, the limit as x approaches 10 is 10+10=20, which is the value that removes the discontinuity. A tempting distractor like 5 might be the originally given value of v(10), but it does not match the limit. To remove removable discontinuities in rational functions, factor and cancel common terms, then evaluate the simplified function at the point.