
Re: Sequence limit
Posted:
Oct 4, 2013 7:12 PM


On Friday, October 4, 2013 3:22:25 PM UTC5, Bart Goddard wrote: > 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."
I am sorry if my being instructor disturbs you. However, if error/s in _my solution_ disturb you, show me at which one of the 4 steps you found an error and I will help you.
> Note that in the real variable > > case, the value of sin x^(1/x) is zero infinitely often.
If you plot a simple graph of y= sin x^(1/x), you won't be able to say that "the value of sin x^(1/x) is zero infinitely often".
> > So the that limit can't be 1.
This is your conclusion based on wrong assumption that "the value of sin x^(1/x) is zero infinitely often". I don't see any logic in it.
Lastly, you mentioned in your original post that "calculus techniques imply the limit evaluates to 1", which specific technique led you to accept that the limit was 1. May be your solution method/technique will lead me to see where your conclusion is coming from.
Best regards.
Mohan Pawar Online Instructor, Maths/Physics MP Classes LLC  US Central Time: 6:12 PM 10/4/2013 > > > > B.

