كم سؤال فالهومورك

91girl · 05-06-2010, 03:14 PM

همسو..

الله يسعدك يارب ممكن تحللي عل الفقرة يارب

91girl · 05-06-2010, 03:23 PM

up uo

الشيمآء · 05-06-2010, 03:28 PM

lim (1-cos(5x))/25x^2 (when x --> 0 ) = 0/0
lim (1-cos(5x))/25x^2 (when x --> 0 ) = lim (5sin(5x))/50x (when x -->0)
= 0/0
lim (5sin(5x))/50x (when x -->0) = lim (25cos(5x))/50 (when x--> 0)
= 25 / 50
= 1/2

HMIS · 05-06-2010, 03:38 PM

المشاركة الأصلية كتبت بواسطة الشيمآء

lim (1-cos(5x))/25x^2 (when x --> 0 ) = 0/0
lim (1-cos(5x))/25x^2 (when x --> 0 ) = lim (5sin(5x))/50x (when x -->0)
= 0/0
lim (5sin(5x))/50x (when x -->0) = lim (25cos(5x))/50 (when x--> 0)
= 25 / 50
= 1/2

yes, shaima is right

^_^