Find the values of p for which the limit exists

[frenchkiki]

Homework Statement

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

Homework Equations

None

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 x_n=1/nπ and y_n=1/(2n + 1/2)π
f(x_n)= sin (1/x_n) = sin (nπ) n=1,2,...,n
lim (n-->∞) x_n = 0
lim f(x_n) = 0

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

p<0

lim (x-->0+) x^p sin (1/x)
f(x_n) = (1/nπ)^p sin (nπ)
f(y_n) = (1/(2n + 1/2)π)^p sin ((2n + 1/2)π)

lim (n-->∞) x_n=0
lim (n-->∞) f(x_n)=0

We have lim (n-->∞) x_n=lim (n-->∞) f(x_n)=0 so the lim (x-->0+) x^p sin (1/x)=0

Not too sure about this conclusion either...

p>0

Using the squeezing theorem:

-1<sin(1/x)<1
-x^p<x^psin(1/x)<x^p

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

Then, lim (x-->0+) x^p=lim (x-->0+) -x^p=lim (x-->0+) -x^psin(1/x)=0

Thanks in advance!

 

[SammyS]

...

p=0

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

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), ...

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

 

[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?

 

[SammyS]

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?

It's the sequence f(x_n) that's of interest here.

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

 

[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

 

[SammyS]

It's the sequence f(x_n) that's of interest here.

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

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

Of course (x_n)⁰ = 1 .

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

 

[frenchkiki]

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

Of course (x_n)⁰ = 1 .

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

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

 

[HallsofIvy]

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

 

[frenchkiki]

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

 

[SammyS]

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

 

[frenchkiki]

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

then lim (n→∞) x_n = 0 ≠ 1 = lim (n→∞) y_n.

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

Thank you both for your help.

 

[HallsofIvy]

Yes, that was exactly what SammyS said.