Find the values of p for which the limit exists

1. Oct 19, 2011

frenchkiki

1. The problem statement, all variables and given/known data

Find the values of p for which the limit exists: lim (x-->0+) xp sin (1/x)

2. Relevant equations

None

3. The attempt at a solution

Alright so I've broken it down to 3 cases:

p=0

then we have lim (x-->0+) x0sin (1/x) = lim (x-->0+) sin (1/x)
Let xn=1/nπ and yn=1/(2n + 1/2)π
f(xn)= sin (1/xn) = sin (nπ) n=1,2,...,n
lim (n-->∞) xn = 0
lim f(xn) = 0

I'm not too sure how to deduce that the limit doesn't exist here.

p<0

lim (x-->0+) xp sin (1/x)
f(xn) = (1/nπ)p sin (nπ)
f(yn) = (1/(2n + 1/2)π)p sin ((2n + 1/2)π)

lim (n-->∞) xn=0
lim (n-->∞) f(xn)=0

We have lim (n-->∞) xn=lim (n-->∞) f(xn)=0 so the lim (x-->0+) xp sin (1/x)=0

p>0

Using the squeezing theorem:

-1<sin(1/x)<1
-xp<xpsin(1/x)<xp

So if we have:
lim (x-->0+) xp=0
lim (x-->0+) -xp=0

Then, lim (x-->0+) xp=lim (x-->0+) -xp=lim (x-->0+) -xpsin(1/x)=0

2. Oct 19, 2011

SammyS

Staff Emeritus
Instead of the sequence used here, pick one that has odd number multiples of 1/(π/2), i.e., 1/(π/2), 3/(π/2), 5/(π/2), ...

3. Oct 19, 2011

frenchkiki

OK then I get lim (n-->inf) x_n=+inf and
lim (n-->inf) f(x_n) =0. Is this enough to show that lim (x-->0+) sin (1/x) does not exist?

4. Oct 19, 2011

SammyS

Staff Emeritus
It's the sequence f(xn) that's of interest here.

So, if $\displaystyle x_n=\frac{1}{(2n-1)\frac{\pi}{2}}\,,$ what is f(xn) ?

Last edited: Oct 19, 2011
5. Oct 19, 2011

frenchkiki

f(x_n)=sin (pi n I pi/2), then I get lim (n-->inf) f(x_n)=+inf and lim (n-->inf) x_n =0

6. Oct 19, 2011

SammyS

Staff Emeritus
First look at 1/xn = (2n - 1)(π/2) = π/2, 3π/2, 5π/2, ... , for n = 1, 2, 3, ...

Of course (xn)0 = 1 .

So f(xx) = sin(π/2), sin(3π/2), sin(5π/2), ... = ? ? ?

7. Oct 19, 2011

frenchkiki

so f(xn) is not defined because it oscillates between 1 and -1?

8. Oct 19, 2011

HallsofIvy

The sequence is certainly defined. It is the limit of the sequence that is not defined.

9. Oct 19, 2011

frenchkiki

What is unclear to me is how to use the sequence to prove that the limit doesn't exist.

10. Oct 19, 2011

SammyS

Staff Emeritus
Break the given sequence up into two sub-sequences; one with the odd terms, the other with the even terms.

11. Oct 21, 2011

frenchkiki

A brief visit at my professor's office clarified the problem. You pick two sequences xn and yn tending to 0. In the case p=0 case f(xn)=0 f(yn)=1 with xn=1/nπ and yn=1/(2n + 1/2)π.

then lim (n→∞) xn = 0 ≠ 1 = lim (n→∞) yn.

Therefore lim (x→0+) xp sin (1/x) does not exist for p=0.

Thank you both for your help.

12. Oct 21, 2011

HallsofIvy

Yes, that was exactly what SammyS said.