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# Note for MATH 263A with Professor Holston at Ohio-group work 1 sol

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COURSE
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TYPE
Class Notes
PAGES
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WORDS
KARMA
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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Ohio University taught by a professor in Fall. Since its upload, it has received 13 views.

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Date Created: 02/06/15
Group Work 1 Solutions 2 lim t70 1 1 t t2 1 Solution This is indeterminate of the form 00 7 00 i 1 t 1 1 1 hm 7 7 1m 7 1 t70 t t1 tt1 t70tl 2 3 hm W 1700 27273 Solution This is indeterminate of the form g oo 2 2 3 4 1 2 2 3 4 2 limlt 95 M lim h M 2 700 272z73 122 z7ltgtltgt172732 72 4 lim z 172 2z74 i i i i 0 Solutlon Thls 1s 1ndeterm1nate of the form 7 72 x72 1 0 72 72 Noticethat lim 7and lim lim 7 hm 7 712 21x721 712 2272 2 DNE 12x 74 72 212 7 2 1 777 and so lim 2 72 72 7 5 Find all points of discontinuity for f tan 1 if z lt 1 fz 227 if1 z 2 4 xgt2 273m Solution The functions tan 1 and 2 7 z are both continuous7 but the function 22 is discontinuous at z 03 Since this particular piece is only used for z gt 27 z 0 is NOT a discontinuity for f The only other potential discontinuities are x 12 Note lirn f lirn tan 1 E 171 171 and lirn f lirn2 7 s 0 Hence lirnfz DNE7 and 1 is a point 71 71 z7 of discontinuity However7 lirgii f lirgijsz 7 s 2 and lirgiJr f 2 7 3w 27 so f 27 and f2 27 so f lS continuous at 2 Thus the points of discontinuity are x 17 3 6 Find the values of the constants A and B which make the function con tinuous at 2 Az76 i m 1fz7 2 n B x2 Solution When trying to plug in z 2 into the top piece7 we get MT G Note that if 2A 7 6 31 07 then ling 92 DNE7 and 9 would not be continuous at 2 Thus we must have 2A 7 6 07 ie A 3 Thus 3 7 6 3 7 2 3 lirngx lirn z irn lirn 7 For g to be 172 72 2 7 4 72 7 2 z 2 z72 z 2 4 3 continuous at 27 we need ling 92 927 where 92 B Thus B

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