Question #a2bce

1 Answer
Jan 18, 2017

The limit:

lim_(x->oo) xsinx is indeterminate.

Explanation:

For every integer N>0 consider points:

x_N = Npi-pi/2

they divide the interval [pi/2,+oo) in a set of contiguos intervals such that every point x> pi/2 falls in one of the intervals:

I_N= [x_N, x_(N+1)) = [Npi-pi/2, Npi+pi/2)

at the boundaries of every interval we have:

For N even:

sin (x_N) = sin(2Kpi-pi/2) = sin(-pi/2) = -1
sin (x_(N+1)) = sin(2Kpi-pi/2+pi) = sin(pi/2) = 1

For N odd:

sin(x_N) = sin((2K+1)pi-pi/2) = sin(pi/2) = 1
sin (x_(N+1)) = sin((2K+1)pi-pi/2+pi) = sin((3pi)/2) = -1

In either case for the theorem of intermediate values, as f(x) = xsinx is continuous in the intervals I_N, it assumes in the interval all the possible values between -x_N and x_N.

This means that given any number L and any numbers M,Q > 0 we can find a point bar x > M such that:

abs (f(bar x) - L) > Q

which is in contradiction with f(x) converging to any limit finite or infinite.

graph{xsinx [-191.3, 212.9, -100.5, 101.6]}