10 points Compute the following limits a lim x sin 3 x 1 cos 2 x e x 3 1 6 sin

Thomas' Calculus: Early Transcendentals

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(10 points) Compute the following limits: (a) lim x 0 sin 3 x (1 - cos 2 x ) e x 3 - 1 = 6 sin 3 x = (3 x ) - (3 x ) 3 3! + (3 x ) 5 5! - · · · cos 2 x = 1 - (2 x ) 2 2! + (2 x ) 4 4! + · · · e x 3 = 1 + x 3 + x 6 2! + · · · As x goes to zero, the higher powers of x die out much more quickly. So the lowest power of x dominates. Thus sin 3 x 3 x , 1 - cos 2 x 4 x 2 / 2 = 2 x 2 and e x 3 - 1 x 3 as x goes to zero. Thus sin 3 x (1 - cos 2 x ) e x 3 - 1 (3 x )(2 x 2 ) x 3 = 6 as x 0 (b) lim x →∞ x 2 e - 1 /x 2 - 1 = - 1 . The Taylor series for e u converges to e u for every real number u . So in particular if u = - 1 /x 2 then e - 1 /x 2 = 1 - 1 x 2 + 1 2 x 4 - 1 3! x 6 + · · ·
Thus x 2 ( e - 1 /x 2 - 1) = x 2 ( - 1 x 2 + 1 2 x 4 - 1 6 x 6 + · · · ) = - 1 + 1 2 x 2 - 1 6 x 4 + · · · As x goes to all the terms except the first die out, so the limit is - 1.

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