Question

For \(p\in Q\) define \(f: R\rightarrow R\) by \(f(x)=[|x|^p sin(\frac{1}{x}) ~~x\not = 0\);\(0~~x=0\). For which values of p is f continuous at 0?

Answer 1

@ganeshie8

Answer 2

@SolomonZelman , I am not sure if you know this, but could you look at it for me?

Answer 3

@freckles do you mind taking a look?

Answer 4

I think it is continuous for any value of p>1...

Answer 5

but I don't know how to prove it

Answer 6

We want f to be continuous at x=0...
we need to have this:
\[\lim_{x \rightarrow 0}|x|^psin(\frac{1}{x})=0\]
Thinking right now

Answer 7

oh, it should include 1. I was thinking differentiable(2nd part of the question)

Answer 8

We know \[\lim_{x \rightarrow 0} x \sin(\frac{1}{x})=0 \text{ by squeeze theroem }\]
-1<sin(1/x)<1
-x<x sin(1/x)<x
and -x and x go to 0 so x sin(1/x) goes to 0

Answer 9

agreed, same for integers above 1

Answer 10

and yeah I didn't put the equal signs there (just the inequalities) because I'm being lazy

Answer 11

Oh true

Answer 12

If you can use squeeze theorem then we are done right? \[-x^n \le \sin(\frac{1}{x} )\le x^{n} \text{ for rational } n \ge 1 \]
we have that the thingy in between goes 0 0 since the thingys on the outside go to zero

Answer 13

it's the rationals I do not know how to prove. I rewrote p as \(\frac{m}{n}\) then changed \(|x|^\frac{m}{n}=(|x|^m)^\frac{1}{n}\) but I'm not sure that helps

Answer 14

can we assume?

Answer 15

and yes, we have squeeze up our sleeve

Answer 16

i wonder if we consider rationals between 0 and 1

Answer 17

so, jsut cause I'm now thinking about it, what about for 0<p<1

Answer 18

like -x^(1/2) and x^(1/2) also go to zero

Answer 19

as x goes to zero

Answer 20

roots are continuous, no?

Answer 21

oh but not on R only R+

Answer 22

but we have an abs. value so yes, they end up eing continuous

Answer 23

so if we had \[f(x)=x^n \sin(\frac{1}{x}), x \neq 0 \text{ and } 0 ,x =0\]
We have by squeeze theorem that when n>0 that \[\lim_{x \rightarrow 0}f(x)=0\]
do you agree with this?

Answer 24

The only think I' haven't touched on is the | | part
but that shouldn't make much of a difference

Answer 25

i think

Answer 26

not in the case of positive exp. No, no difference

Answer 27

but yes, I agree

Answer 28

|| doesn't effect cont. only differentiability

Answer 29

that is true because |x|->0 as x->0

Answer 30

any |x|^n->0 as x->0 as along as n>0

Answer 31

and not any*

Answer 32

? not any? every exponent goes to 0 when x goes to zero.

Answer 33

every positive exp*

Answer 34

I was just saying i didn't mean to write any
I meant to write and

Answer 35

oh ok, you typed any again

Answer 36

So now, do you agree that for p>1 it is differentiable?

Answer 37

at zero

Answer 38

\[f(x)=x^n \sin(\frac{1}{x}) \text{ if } x>0 , \text{ and } (-x)^n \sin(\frac{1}{x}) \text{ if } x<0,\text{ and } 0 \text{ if } x=0\]

Answer 39

I was going to look at the derivative of this
and I think you are right
but I have to go right now
i have a meeting at 1 oclock
sorry

Answer 40

it's cool thanks for the help!

Answer 41

we don't care about p<=0
because we already said it wasn't continuous on p<=0
so it certainly won't be differentiable there
So we need to look at p>0 only \[f'(x)=nx^{n-1}\sin(\frac{1}{x})+x^n \cdot \frac{-1}{x^2}\cos(\frac{1}{x}) \text{ for } x>0 \\ f'(x)=nx^{n-1}\sin(\frac{1}{x})-x^{n-2}\cos(\frac{1}{x}) \text{ for } x>0 \\ f'(x)=x^{n-2} \cdot [n x \sin(\frac{1}{x})-\cos(\frac{1}{x})] \text{ for } x>0 \\ \text{ and } \\ f'(x)=-n(-x)^{n-1}\sin(\frac{1}{x})+(-x)^{n} \cdot \frac{-1}{x^2}\cos(\frac{1}{x}) \text{ for } x<0 \\ f'(x)=-(-x)^{n-2}[(-x)\sin(\frac{1}{x})+\cos(\frac{1}{x})] \text{ for } x<0\]

Answer 42

we should check to see if the function is actually differentiable at x=0
that is we need to check to see
\[\lim_{x \rightarrow 0^+}f'(x)=\lim_{x \rightarrow 0^-}f'(x)=0\]
so f'(x)=0 when x=0
so actually n might have to be greater than 2

Answer 43

And I say n might have to be greater than 2 because of x^(n-2) factor