
Re: Sequence limit
Posted:
Oct 4, 2013 4:22 PM


Mohan Pawar <GoogleOnly@mpclasses.com> wrote in news:02fcfd2855fd48ff8dcfbda8e1753bbb@googlegroups.com:
> On Thursday, October 3, 2013 1:01:06 PM UTC5, Bart Goddard wrote: >> This question from a colleague: >> >> >> >> What is lim_{n > oo} sin n^(1/n) >> >> >> >> where n runs through the positive integers. >> >> >> >> Calculus techniques imply the answer is 1. >> >> But the same techniques imply the answer is 1 >> >> if n is changed to x, a real variable, and that >> >> is not the case, since sin x =0 infinitely often. >> >> >> >> Anyone wrestled with the subtlies of this problem? >> > > In fact there is no subtlety if you see why the limit is 1 in the > implementation of calculus techniques mentioned above for both cases > i.e. when > > (1) variable is n =1, 2, 3
> > (2) variable is x is any real number > > I will consider first the general case of x as real that can be used > to get specific case when x is n > > Solution: In given lim x > inf. sin x^(1/x) > > 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) > > 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. Now if m is replaced by natural > number n, the situation does not change sin (1/n)will still be > within 0 to 1 and limit will evaluate to due to zero in exponent. > > Mohan Pawar > Online Instructor, Maths/Physics > MP Classes LLC >  > US Central Time: 1:53 PM 10/4/2013 >
The only thing that disturbs me about this is that you're an "Instructor." Note that in the real variable case, the value of sin x^(1/x) is zero infinitely often. So the that limit can't be 1.
B.

