فارسی · Curriculum · Week 6 guide · Back to exam
- 20 points. Factor as (x − 6)(x + 6) (6), then cancel only for x ≠ 6 (5). The limit is 12 (5). An assigned value of 20 at x = 6 does not change surrounding values, so the limit stays 12 (4).
- 20 points. Multiply by the conjugate √(x + 16) + 4 (6). The resulting numerator is x, which cancels for x ≠ 0 (6). The new denominator approaches 8 and is nonzero (4), so the limit is 1/8 (4).
- 10 points. On the left the quotient is −1 and on the right it is +1 (3 each). Since the one-sided limits differ, there is no two-sided limit (4).
- 30 points. Setting VB to zero gives vA = 10/2 = 5 V (6). Setting VA to zero gives vB = −8/4 = −2 V (6). Add to get v = 3 V (6). The outgoing currents are −7/4 = −1.75 mA, 11/8 = 1.375 mA, and 3/8 = 0.375 mA (3 each). Their sum is zero (3).
- 20 points. The combined conductance with both voltage sources inactive is 1/2 mS, so the injection contributes 2 V (5). The combined node voltage is 5 V (4). Load power is 25/8 = 3.125 mW (5). Power depends on the square of the combined voltage, so summing squares omits the cross terms (6).