Ban. Func. QB 18

-1

wast da ans bhai jadi batao........

thn i'll post da workinnnnnnnnnn

e^1/e
whats the answer??????

mani....

2 verify ur ans.........

take a value of the order 10^-4 wich is very clos to 0, so u'll get an idea wethr ur ans is rite or not....

maine bhi..... obviously workin se hi ans aata hai, par this is a meth 2 re-chk ur ans!!!
exam me not useful par here u can do dat............

how will u take log directly as the terms r in subtraction......

I'll post da workin once i get da affirmation regardi my ans's validity.........

limx^xx=M
=>logM =lim x^xlogx
=>log(log M)=lim (xlogx +log(logx))
=> log(log M)=lim xlogx +lim log(logx)
=> log(log M)=0+lim log(logx)
yahaan tak to theek hai.....
now maybe => M=0 =>lim f(x)=1

I'll tell my meth........

let t = x^x

y=x^t - t;

logt = xlogx ............(lim x-->0)
= logx/(1/x)......... not defined / ∞

differentiating...........

logt = (1/x)/(-1/x²)
logt = 0

ther4 t=1

y= x^t - t;

applying limits.......

y = 0^1 - 1;
y = 0 - 1

y =-1 [1] [1]

first part to hum dono ka same hi hai bhai.......

tapan seems correct to me!

no chem....

chk properly that wont b the case...........

and neways 0^0 ≠0 as it isnt defined