# 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.