1) If f(x) = lim n→ ∞ n ( (x) 1/n -1 )
for x>0, y>0 , then show that f(xy) = f(x) + f(y)
2) Solve lim x→0 [1+log cos 2 x/2 cosx] 2
111) f(xy) = lim n→ ∞ n [(xy) 1/n -1]
= lim n→∞ n {y 1/n (x 1/n -1) + (y 1/n -1 )}
= lim n→∞ y 1/n lim n→∞ n (x 1/n -1 ) + lim n→∞ n (y 1/n -1 )
= y 0 . f(x) + f(y)
= f(x) + f(y)
Therefore, thus proved.
12) separate d limits. independently solve lim ( x→0) 1 + logcos2x/2 cosx.
to solve it separate the limits again. u'll get : 1 + ln cosx / ln cos2x/2 .
now, multiply and divide cos x / cos2x/2 and u'll get in the formlim(x→0) ln (1+x) / x = 1
now, multiply the independent ones. u'll get answer as 4.
13571) f(x)=limn→∞(x1/n-1)/(1/n)=limt→0(xt-1)/(t)=logx
f(x)=logx
so f(xy)=f(x)+f(y)
