Math 165 - Quiz 2B, trig limits - solutions Problem 1 Find the indicated limit. t cos(2t) + sin(5t) t→0 3t L = lim Solution Split it up! First, t cos(2t) cos(2t) = , 3t 3 and the limit of this for t → 0 is 1/3, simply by substitution. Second, lim t→0 sin(5t) = 1. 5t Bring this into the second half of the problem (if you like, you could multiply both sides in the preceding equation by 5/3, or you do it in the next equation): 1 sin(5t) + lim 3 t→0 3t sin(5t) 5 1 + lim · = 3 t→0 5t 3) 1 5 = + = 2. 3 3 L = Make sure you never divide by zero in problems like this! this is probably the fastest way to lose lots of points. Problem 2 Which choice of the constant a will make the function f (x) below continuous everywhere? 4 + x2 for x < 1 f (x) = ax − 1 for x ≥ 1 Solution The function f (x) is continuous everywhere except perhaps not at x = 1, because of the two different definitions meeting there. So we need to ensure lim− f (x) = f (1) = lim+ f (x). x→1 x→1 2 For the left-hand limit, use 4 + x and substitute x = 1. For the right-hand limit, use ax − 1 and substitute x = 1 again. So 4+1=a·1−1 and therefore a = 6 is the only choice. Imagine that you are supposed to attach a metal rod to a curved machine part (eg arm of a forklift). The curved part has a surface with equation y = 4 + x2 , for x < 1, and your robot raises the metal rod with equation y = ax − 1 (starting at a = 5, 5.1, 5.2, . . . ) until it touches the forklift arm. If you don’t want trial and error, you need to figure out the value of a when that happens.