كم سؤال فالهومورك

[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

[HMIS]

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

^_^