10FallHwk03_Soln

10FallHwk03_Soln - Math 2413 Homework Set 3 Solutions z ( 6...

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Math 2413 Homework Set 3 – Solutions 10 Points 2. (2 pts) ( )( ) ( ) ( ) 2 6 6 1 0 10 54 3 42 6 li m li m li m lim 1 0 6 1 0 6 6 66 z z zz z z z z ¥ ¥ ¥ fi¥ -- -+ - - == = = = -¥ ++ 5. (2 pts) ( ) ( ) ( ) ( ) 2 2 22 27 2 7 2 2 77 1 4 14 2 14 li m li m li m Assume 0 because so 6 67 6 14 14 14 li m lim 6 y yy y y y y y y y y y y y y ¥ ¥ fi¥ ¥ fi¥ + = = > ¥= + + + + + = For the second limit the work will be identical until we get rid of the absolute value. In this case, we can assume that 0 y < because y fi -¥ and so =- . So, picking up we get, ( ) 2 2 2 14 14 2 1 4 14 li m li m lim 6 y y y y y y y ¥ ¥ fi¥ + + + = = - + + 7. (2 pts) Not much to do here. The function is continuous because it is a difference/product of continuous functions. Now all we need to do is evaluate the function at the two points. ( ) ( ) 1 5 11.792 3 7 4.0547 ff - = - So, we can see that ( ) ( ) 7 4.054 7 0 1 5 11.7923 - = - <<-= and so, by the IVT there is a number c such that 1 57 c - < <- such that ( ) 0 fc = . Or in other words, c is a root of the function. For the sake of completeness we can use a computer to see that 11.48494037 x is a root of the function. 10. (2 pts) () ( ) () ( ) ( ) ( ) ( ) ( ) ( ) 2 2 00 2 2 0 2 3 4 34 li m lim 3 2 4 lim 63 li m lim 6316 1 61 hh h x h x h xx fx h fx x x h h x h h x hhh x h x f h fifi + -+ + - +- ¢ -+ + - = ¢ = = + -= - & 11. (2 pts)
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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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10FallHwk03_Soln - Math 2413 Homework Set 3 Solutions z ( 6...

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