Lim Sin.Log x2
x→0
Find the value using L hospital rule.
51Not Possible.
Himanshu Giria hmmmm L hospital is applicable for 0/0 and infinity by infinity Himanshu Giria hmmmm L hospital is applicable for 0/0 and infinity by infinity 
Karan Matalia you have to convert it to 0/0 form or something like that
591Actually i think the limit is not defined..as logx2→∞ as x→0 so the limit is sin(∞) and it can have any value b/w -1 and 1..u can't exactly say
As for the method to use in some other problem,
Let L= lim x→0 Sin.Logx2
L = lim x→0 xSin.Logx2 x
Differentiate the N & D
L= lim x→0 (SinLogx2+12CosLogx2)
L=L+lim x→0 12CosLogx2
so lim x→0 CosLogx2= 0
from here lim x→0 Sin.Logx2 = ±1 ( from this method too 2 values are coming which is absurd)