10FallHwk02_Soln

# 10FallHwk02_Soln - Math 2413 1. (2 pts) (a) Homework Set 2...

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Math 2413 Homework Set 2 – Solutions 10 Points 1. (2 pts) (a) ( ) ( ) ( ) () () () 11 1 1 2 li m 2 li m1 m doesn't exist b/c li m lim xx x f f x fx f x f x -+ - ﬁ- - ﬁ- - = == =„ (b) ( ) ( ) ( ) ( ) 2 22 23 m 1 li m 1 li x f f x f x ﬁﬁ = - = - = - =- (c) ( ) ( ) ( ) ( ) 3 33 3 doesn't exis t li m 4 m 4 li m4 x f f x f x = = == 3. (2 pts) ( ) ( )( ) 2 2 4 44 4 4 li m li m lim 3 2 8 4 7 7 1 1 11 x x x x x - - ﬁ- + +- = = - - + - -- 5. (2 pts) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 7 77 7 2 2 5 32 2 53 4 2 3 21 li m li m lim 7 2 2 5 3 7 2 2 5 3 7 2 2 lim 4 2 2 y yy y y y y y y y y - - - - + - - + - - +- 6. (2 pts for b ONLY) (a) In this case we can use the first formula because x is completely inside this region. ( ) ( ) 1 1 11 li m lim 4 1 43 g - ﬁ- = + =- (b) Here we will need to look at the two one-sided limits because 6 x = is the “cut-off” point. ( ) ( ) () () () 66 li m lim 41 25 li m lim li m lim x gx ++ = += The two one-sided limits are not the same and so ( ) 6 lim x doesn’t exist. 7. (2 pts) Note that we can’t just cancel the h ’s since once is inside the absolute value bars. So, we’ll use the hint and recall that, if 0 if 0 hh h & = ± -< ² With this we can do the two one sided limits to eliminate the absolute value bars. 000 li m li m li m 1 1 because 0 in this case hhh h h h - ﬁﬁﬁ - = = - = 0 00 li m li m lim 1 1 because 0 in this case h hh h h

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## This note was uploaded on 11/13/2010 for the course MA 141 taught by Professor Wears during the Spring '07 term at N.C. State.

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10FallHwk02_Soln - Math 2413 1. (2 pts) (a) Homework Set 2...

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