X 0 x x 0 0 x 0 0 0 x0 0 x x0 x x0 now

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Unformatted text preview: ction Take the ln of 0 0 1 both sides 0 ,1 ,1 Case 0 0 0 01 0 0 0 0 11 0 , A,ction , 0 ,011 1, 1 1 1,1 We found 0 ake the ln of T ,1 ,1 this in an 0 01 both sides earlier L’hopital’s Rule slide x0 0 lim+ y ln of both sides x lim xxx0(0 == lim y 0 limx ( 0 ) ) lim+ x (lim!x+lim+)y0 ) Then take 0 x + 0 (0 (0 = = x!+ y ) lim xx = lim + x!0 x!0 x!0+ + !0 0 x!0 x!0x!0+ x x!0 0 ,1 ,1 0 0 ,1 ,1 0 1 x lim+ y ) = ) lim+ ln(x )x) x)lim0 x ln(x) = 0 y =limyln(ln(xlimlim lim+xx== lim0+x ln(x) = 0 lim+=)) lim!0 ln(x == x! x ln(x) = 0 = ln( 0ln(y = x =ln(xx ) ln( lim ln(x ) ln(+ x!x! ) 0 +x + x! 0+ ! 0 x ! 0+ 0+ 0 x!0++ ! 0+ x x!0 ! x x!0 Now take exp of both sides lim 0 lim0!= ln(y ) x!0+ ln(y ) = e0 0 1 + ) ) xlim+)ln(y )e= ) 0 expe = lim= 1yy )) = e0 = 1 = 0 ln(y )0==x =) exp ( e ln( )) = == 1 =) lim+ = 0 ln(0 ( lim+ ln(y !y ln(y ) = 0 =) exp ( xlim+ + ln( )) = e = 1 + !0 x x0 !!0 x!0 ln(y ) ln(y ) lim e ln(y ) = lim y = lim xx = 1 e = ln(y+ y ==lim+ + x = 1lim+ xx = 1 lim x ln(e x lim+ 0 )) = x!lim+ y lim0 +xx = 1 +y x!x! 0 x!0y = lim x = 1 x! limx e e lim+ y lim +!0 = lim0 0 = x!+ x! 0 x = x! 0 x ! 0+ x!0 x ! 0+ x!0 x ! 0+ Optimization Problems ¤༊ PRACTICE, PRACTICE, PRACTICE!!! lim+ ln(y ) = lim+ ln(x ) = lim+ x ln(x) = 0 x!0 x!0 x!0 ln(y ) x limln(y lim+ + e ) = 0 = lim( lim= y )) = e0 = = =) exp + y + ln( lim x 1 x!0Summation !0 x!0 x x ! 0+ x !0 1 lim+ eln(y) = lim+ y = lim+ xx = 1 Two strategies: x!0 x! 0 x!0 ✓ X ◆ 1. Write out100 first couple terms and notice a the pattern (maybe the middle ones cancel out) 1 100 X ✓1 i i=1 i=1 i 1 ◆ 1 i+1 i+1 2. Simplify down to the summation formula’s we n n X X Xn know: nn n X X2 , X i2 i3 3 i, i=1 i ,=1 i i=1 i i=1 i, i=1 i=1 Integration Think of an integral as the area under a curve....
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This document was uploaded on 03/04/2014 for the course MATH 1A03 at McMaster University.

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