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MATH 111 Final: MATH 111 NJIT M111 FinalF14

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turquoisegnu128

15 Feb 2019

School

New Jersey Institute of Technology

Department

Mathematical Sciences

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Professor

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All work must be shown in order to receive full credit. The use of books, notes, calcu- lators, or any other external sources of information is not allowed during this examination. Evaluate the following limits: x2 4: lim x 2. X2 + 5 3 sin(x) x 0+ 1 lim c. 1 x2 arcsin(x) a. lim x 1 : lim x 1 sin( x) ln(x) Find y (x) for the following: y = x3 arcsin(5x, y = ln(y cos(x)) Find y (x) for the following: y = sin( x) 1 x ln(x) dx x: z 1, z sin(x) 4 + cos(x) dx b. lim x 0+ x 1 ln(x) b. lim x sin(x)csc(x: y = x2x + x2, y = z 10 x3 sin(t2) dt, z 4. Evaluate the following integrals: z x + 2 x2 + 1 dx, z x sin x2. 4 dx: z x3px2 + 4 dx.

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Related Questions

3. (21 pts) Evaluate the limit or say that the limit does not exist. If a limit does not exist, explain why. Show your work. You may NOT use a table of values, a graph, or L'Hospitals's Rule to justify your answer. (Note: Possible answers include + or -..) (a) (7 pts) x2 - 2x - 3 - lim x+3 Vx+1 - 2 x2 - 2x - 3 Vx+ 1 + 2 lim x+3 Vx+1 -2 Vx+1 +2 (x - 3) (x + 1) (Vx+ = lim x+3 x + 1 - 4 = lim x+3 (x - 3) (x + 1) (Vx+1 +2). -- x-3 = lin (x + 1) (Vx+1 +2) = (3 + 2) (V3+1 +2) = = 4 (V4 +2) = 16 (b) (7 pts) lim xsin (In x) X >0+ NOTE: We cannot apply the Product Law, since lim sin (1n x) DOES NOT EXIST. On the other hand, -1s sin (in x) s 1 and , therefore, -xs x sin (in x) sx, (since x > 0). Since lim X = lim (-X) = 0, x0+ X0 the inequality above implies that lim xsin (in x) = 0, by the Squeeze Theorem . x0+ EXPLANATION : Since lim ln x = -0, x →0+ sin (lnx) oscillates between 1 and -1 as x + 0*, and, therefore lim sin (In x) DOES NOT EXIST. x >0+ (c) (7 pts) in x+2- (lnx) +In 20 -= -00 (x - 2) 70-

Solve all please.

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