# Continuety parameter question

1. Jan 18, 2009

### transgalactic

i tried to solve it
but i didnt get any parameter value
??

2. Jan 18, 2009

### BobMonahon

Hi,
Re: part A: I think this is a "setup" to make you think the answer is easy. Yes, it appears that: for m =1,2,3,4 ... etc, the function is continuous at x=0. Your logic looks good to me.

However, the sin(1/x) function oscillates infinitely fast as x --> 0, so I think you'll find the answers to B and C to be surprising.

3. Jan 18, 2009

### transgalactic

how to solve it?

4. Jan 18, 2009

### BobMonahon

Can you differentiate f(x) "formally" ? I get this:

f'(x)= mxm-1sin(1/x) + xm-1cos(1/x)(-1)x-2

For the case of m=1, I get:

f'(x) = sin(1/x) - cos(1/x)/x2

The first term doesn't approach anything as x--> 0. It oscillates infinitely fast between +1 and -1. No limit, therefore not continuous.

Also, in second term, cos(1/x) -->1, but 1/x2 --> infinity, so again no limit.

Your assignment, Mr. Phelps, should you choose to accept it ... :) ... is to find an m > 0 where f'(0) is continuous. See if you can find one.

5. Jan 19, 2009

### transgalactic

i got a solution to A
and it says

if m=0 then
$x_n = \frac{1}{(2n)\pi}$
i dont know how they came up with this Xn
and what i need to do with it
??

6. Jan 19, 2009

### BobMonahon

Hello,
Oops, I missed the m > OR = 0 part.
For f(x) = xmsin(1/x)
If m=0, then f(x) = sin(1/x)
So, for m=0, your logic for testing continuity of f(x) at zero breaks down. Sin(1/x) is not continuous at 0.

Now, regarding that "solution" you found that says:
$x_n = \frac{1}{(2n)\pi}$
Hmm. Well, let's look at when sin(1/x) = 0
then 1/x = {$\pi$, 2$\pi$, 3$\pi$, ... } = n $\pi$
So I get: x = $\frac{1}{(n)\pi}$
Different than your solution, but the same idea: as n gets larger, x gets closer to 0, but ... between every point where sin(1/x) = 0, it oscillates up to +1 (or down to -1):

when sin(1/x) = 1 --> 1/x = {$\pi$/2, $2\pi$+$\pi$/2, ...} = $2n\pi$+$\pi$/2 --> x = $\frac{1}{(2n\pi + \pi/2)}$
and when sin(1/x) = -1 --> 1/x = {$3\pi$/2, $2\pi$+$3\pi$/2, ...} = $2n\pi$+$3\pi$/2 --> x = $\frac{1}{(2n\pi + 3\pi/2)}$

So ... as x ->0, it passes through a full oscillation between each successive $x_n = \frac{1}{(2n)\pi}$, taking on values 0, 1, 0, -1, 0 successively between xn and xn+1

That's a long-winded answer. Does it help?

Regards, BobM