Connecting multiple representations of limits is a key skill in AP Calculus BC that requires interpreting tables, graphs, and algebraic forms to determine limit values. The table showing values approaching -1 from the left and 1 from the right indicates the limit does not exist as x approaches 0. The graph with a jump at x=0 represents different one-sided limits. Algebraically, r(x) = |x|/x is -1 for x<0 and 1 for x>0, confirming the limit does not exist. A tempting distractor might be choice A, which incorrectly claims the table approaches 1 from both sides, overlooking the absolute value's effect. To connect representations effectively, always check one-sided limits in functions with absolute values or discontinuities.

Connecting multiple representations of limits involves understanding how algebraic, graphical, and tabular forms convey the same limiting behavior. Algebraically, for q(x) = |x|/x, the left-hand limit is -1 and the right-hand limit is 1, so the overall limit does not exist. Graphically, a jump discontinuity at x = 0 illustrates the differing one-sided limits. Tabular values approach -1 from the left and 1 from the right, confirming the discrepancy. Choice B fails as a tempting distractor by claiming the limit is 0, perhaps averaging the sides incorrectly. A transferable strategy is to evaluate one-sided limits algebraically, identify jumps or asymptotes graphically, and examine directional approaches in tables.

This problem requires connecting multiple representations of limits for s(x) = 1/(x-4). Algebraically, as x approaches 4 from the left, x-4 approaches 0 through negative values, making s(x) approach -∞; from the right, x-4 approaches 0 through positive values, making s(x) approach +∞. A table would show s(3.9) = -10, s(3.99) = -100, s(4.1) = 10, s(4.01) = 100, confirming the one-sided limits approach opposite infinities. Graphically, this creates a vertical asymptote at x=4 with the function going to -∞ on the left and +∞ on the right. Choice B incorrectly claims the limit is 0, perhaps thinking of horizontal asymptotes instead of vertical ones. When one-sided limits approach different infinities, the two-sided limit does not exist, which all three representations must consistently show.

This problem requires connecting multiple representations of limits for a piecewise function u(x). Algebraically, for x<1, u(x) = x+2, so the left-hand limit is 1+2 = 3; for x>1, u(x) = x²+1, so the right-hand limit is 1²+1 = 2; since these differ, lim(x→1) u(x) does not exist. A table would show u(0.9) = 2.9, u(0.99) = 2.99 from the left and u(1.1) = 2.21, u(1.01) = 2.0201 from the right, confirming different one-sided limits. Graphically, this creates open circles at (1,3) from the left piece and (1,2) from the right piece, with a filled dot at (1,3) for the actual value u(1)=3. Choice A incorrectly claims the limit exists and equals 3, considering only the left-hand limit. For piecewise functions, verify that all representations consistently show whether one-sided limits match to determine if the two-sided limit exists.

This problem requires connecting multiple representations of limits for h(x) = sin(x)/x, a famous limit in calculus. Algebraically, using L'Hôpital's rule or the squeeze theorem, we can show that lim(x→0) sin(x)/x = 1. A table of values confirms this: h(0.1) ≈ 0.998, h(0.01) ≈ 0.99998, h(-0.1) ≈ 0.998, h(-0.01) ≈ 0.99998, showing convergence to 1 from both sides. Graphically, this appears as an open circle at (0,1) where the limit occurs, and a filled dot at (0,2) showing the assigned value h(0)=2. Choice A incorrectly claims the limit is 2, confusing the function value h(0)=2 with the actual limit of 1. When analyzing limits, focus on the behavior as x approaches the point, not the value at the point itself.