9A_quiz2_solns

# 9A_quiz2_solns - Calculus 9A Quiz 2 Solutions Spring 2007...

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Calculus 9A - Quiz 2: Solutions Spring 2007 Name: Evaluate the following limits in 1 and 2: 1. lim y 0 sin 3 y 4 y First of all, we recall that, if c is a number, then lim x a cf ( x ) = c lim x a f ( x ); therefore lim y 0 sin 3 y 4 y = lim y 0 1 4 ± sin 3 y y ! = 1 4 lim y 0 sin 3 y y . So now we’ll evaluate lim y 0 sin 3 y y . To do this we’ll use a variable substitution: Let u = 3 y . Then note that u 0 when y 0, and solving for y in terms of u gives us y = 1 3 u . Therefore the limit lim y 0 sin 3 y y = lim u 0 sin u 1 3 u . Let’s evaluate this last limit now, using the fact that we know already that lim x 0 sin x x = 1: lim u 0 sin u 1 3 u = lim u 0 3 ² sin u u ³ = 3 lim u 0 ² sin u u ³ = 3 . Therefore lim y 0 sin 3 y 4 y = 1 4 lim y 0 sin 3 y y = 3 4 lim y 0 sin y y = 3 4 .

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2. lim θ →-∞ cos θ θ For this we use the ‘sandwich’ theorem or ‘squeezing’ theorem: We know that, for all values of θ , we have - 1 cos θ 1 . Now we would like to take the limit as
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9A_quiz2_solns - Calculus 9A Quiz 2 Solutions Spring 2007...

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