Prelim1Solutions

This preview shows pages 1–3. Sign up to view the full content.

Math 1110 Prelim 1 Answers Please ﬁnd below abbreviated answers to the questions on the ﬁrst prelim. On several of the questions, more explanation would be necessary to receive full credit. 1. ( 20 points ) Compute the following limits. (a) ( 5 points ) lim x 1 (3 x 3 - 2 x + 5) (b) ( 5 points ) lim x →∞ 3 x 2 + 10 x - 1 2 x 2 - 25 x + 120 (c) ( 5 points ) lim x →- 2 1 2 + 1 x x 2 - 4 (d) ( 5 points ) lim x 0 6 x 2 cos x sin x sin 2 x Solution (a) lim x 1 (3 x 3 - 2 x + 5) = 3 · 1 3 - 2 · 1 + 5 = 6 (b) ( Please note that in order to have received full credit on this question, you would have had to do polynomial division. ) Since the degree of the numerator and the degree of the denominator are the same, the limit is just the ratio of the leading coeﬃcients. That is, lim x →∞ 3 x 2 + 10 x - 1 2 x 2 - 25 x + 120 = 3 2 (c) lim x →- 2 1 2 + 1 x x 2 - 4 = lim x →- 2 2+ x 2 x ( x + 2)( x - 2) = lim x →- 2 1 2 x ( x - 2) = 1 16 (d) lim x 0 6 x 2 cos x sin x sin 2 x = lim x 0 3 cos x x sin x 2 x sin 2 x = 3 2. ( 15 points ) Let f ( x ) be the function deﬁned below. f ( x ) = x 2 x ≤ - 1 - x - 1 < x < 1 3 x - 3 1 x < 2 5 x = 2 2 x - 1 x > 2 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(a) ( 5 points ) Is f continuous at x
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}