Please help me out someone!
Lim Sin.Log x2
x→0
Find the value using L hospital rule.
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4 Answers
Aditya Agrawal
·2014-04-17 19:02:39
Not Possible.
- Himanshu Giria hmmmm L hospital is applicable for 0/0 and infinity by infinity Upvote·0· Reply ·2014-04-17 20:18:29
- 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
Akshay Ginodia
·2014-04-27 09:06:16
Actually 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)