# B0 x c 3 2 b x c 3 x x c 3 c2 2 b 2

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Unformatted text preview: e number cB) there exists a corresponding number \$  0 such that for all x, x!  x  x! b \$ Ê f(x)  cB. ! b xÄx ! c # # Ê kx b 5k  # Now, 1 (x b 5)  B  ! Í (x b 5)  " ÈB # " B Í kx b 5k  x Ä c& (x b 5) " " ÈB . Choose \$ œ " ÈB . Then 0  kx c (c5)k  \$ Ê " (x b 5)  B so that lim œ _. xÄx # 50. For every real number B  0, we must find a \$  0 such that for all x, 0  kx c (c5)k  \$ Ê # # 2 \$ œ É B , then 0  kx c 3k  \$ Ê # # Now, 2 (x c 3) c2 (x c 3)  cB  0 so that lim # c2 (x c 3)  cB  ! Í B0 Í (x c 3) 2  " B Í (x c 3)#  x Ä \$ (x c 3) c2 2 B 2 Í !  kB c \$k  É B . Choose œ c_. 1 (x b 5) # 49. For every real number cB  0, we must find a \$  0 such that for all x, 0  kx c 3k  \$ Ê # # Ê c" x "  cB so that lim c x œ c_. xÄ! # # c " x  cB  ! Í " x B0 Í x  # " B Í kxk  " ÈB . Choose \$ œ " ÈB , then 0  kxk  \$ Ê kxk  " lx l  B. Now, " Ê &quo...
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## This note was uploaded on 09/20/2010 for the course MATHEMATIC 09991051 taught by Professor Dr.maenshadeed during the Fall '10 term at Norwegian Univ. of Science & Technology.

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