quasi
Posts:
12,018
Registered:
7/15/05


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


quasi wrote: >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)
I meant: lim (m > 0) sin(1/m)^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
I meant: sin(1/m)^m = 0
>hence for those values of m, > > sin(1/m)^(1/m)
I meant: sin(1/m)^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
I meant: sin(1/m)^m = 1
>hence for those values of m, > > sin(1/m)^(1/m)
I meant: sin(1/m)^m
>approaches 1. > >It follows that the limit > > lim (m > 0) sin(1/m)^(1/m)
I meant: lim (m > 0) sin(1/m)^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.
I meant: such that sin(1/m)^m = c.
Ugh.
By obliviously copying and pasting one typo many times, I ended up with a lot of typos.
Sorry for the confusion.
quasi

