quasi
Posts:
10,563
Registered:
7/15/05


Re: Sequence limit
Posted:
Oct 4, 2013 4:42 PM


Mohan Pawar wrote: > >Let > > x=1/m where m is real, inf< x and m <inf. > >=> as x>inf., m>0 >=> lim x > inf. sin x^(1/x) >= lim m > 0 sin (1/m) ^(m) >= 1 (see below why 1)
No, it's not equal to 1.
In fact, The limit
lim (m > 0) sin(1/m)^(1/m)
does not exist.
>Note that the value of sin(1/m) varies from 0 to to 1 >BUT exponent m is guaranteed to be zero as m > 0.
No. The exponent m is only guaranteed to _approach_ 0.
As m approaches 0, there are infinitely many values of m such that sin(1/m) = 0. For those values of m,
sin(1/m)^(1/m) = 0
hence for those values of m,
sin(1/m)^(1/m)
approaches 0.
On the other hand, as m approaches 0, there are infinitely many values of m such that sin(1/m) = 1. For those values of m,
sin(1/m)^(1/m) = 1
hence for those values of m,
sin(1/m)^(1/m)
approaches 1.
It follows that the limit
lim (m > 0) sin(1/m)^(1/m)
does not exist.
In fact, for any real constant c between 0 and 1 inclusive, there exists an infinite sequence of values of m approaching zero such that sin(1/m)^(1/m) = c.
quasi

