1yes ans is b ...but how?
If f(x)=lim (sinx)2n,then f is
n→∞
A) conti. at x=Î /2
B) disconti at x=Î /2
C)disconti at x=Î
D) none of these
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8 Answers
62hmm.. this is sort of simple... (Except that u must have n-> infinity!)
obvious point of contention is Pi/2
b is the correct choice..
can u make some guess why
62see at all points other than pi/2 the value of this function is less than 1
so what happens is that when you take the power to infinity it will tend to zero except at pi/2 where it is 1
so the limit of pi/2+ and pi/2- will be zero .. while at pi/2 it is 1..
hence the result..
1f(x)=lim n->∞ (sinx)^2n = f(x)=lim n->∞ [1- (1-sinx)]^(1/(1-sinx))2n(1-sinx)
now this thing... [1- (1-sinx)](1/(1-sinx))2n(1-sinx) will not be defined for sinx=1 i.e. x=pi/2
hence f(x) is discont at x=pi/2.
can we interpret it in this way??
1oh nooooooo
i am extremely sorry.......
have not seen that its n tends to infinity........ thot x tends to infinity......
sorry :(
62no dear i am afraid not...
the only thing to look at here is that it is some number smaller than 1 raised to the power infinity..
ur solution is not correct...
u have unfortunately complicated a simpler thing...
if u look carefully at the logic i gave..
and if u see ur logic.. u are saying that u divide and multiply by 1-sinx
i can do that for (1/2-sinx) and take 1/2-(1/2-sinx) in lower number.. that logic will work for 1/2 as well..
i dont know how to tell the flaw in ur logic :(