1. ## continuety parameter question..

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

2. Here's a thought: what happens when m=0 ?

3. 0^0 *bounded is an undefined form

4. Originally Posted by transgalactic
0^0 *bounded is an undefined form
Look carefully... the definition $x^m \sin \frac1{x}$ holds only for $x \neq 0$. You cant simply substitute 0 in the place of x to check for continuity.

$f(x) = \begin{cases} \sin \frac1{x} & x\neq 0 \\ 0 & x = 0 \end{cases}$

Is f(x) continuous at x = 0?

(This is a very popular problem. I am sure you would have seen it somewhere in your notes/text)

5. ok so its 1*bounded wich is not defined either

??

6. Originally Posted by transgalactic
ok so its 1*bounded wich is not defined either

??
The function is well defined. The point of my post was to ask if f(x) is continuous at 0. I claim it is not. Can you prove it?

To prove it is discontinuous at 0, observe that sequence $x_n = \frac{2}{(2n+1)\pi}$ converges to 0, but $f(x_n) = (-1)^n$

7. how does it finds the values of the m parameter

8. First prove the following result:

(*) $\lim_{x \to 0} x^m \sin \frac1{x}$ exists and equal to 0 iff $m > 0$.

Now for part (A) we need to find m for which $\lim_{x \to 0} f(x) = f(0) = 0$, Thus we need m for which $\lim_{x \to 0} x^m \sin \frac1{x} = 0$. From the result (*), this happens iff $m > 0$.

For part (B), we need values of 'm' for which $\lim_{x \to 0}\frac{f(x) - f(0)}{x - 0}$ exists. But $\lim_{x \to 0}\frac{f(x) - f(0)}{x - 0} = x^{m-1} \sin \frac1{x}$. Again by the result, the limit exists iff $m-1 > 0 \implies m > 1$

Do part (C) yourself, its instructive

by solving A and B
it is continues on x=0 when m>0
it is differentiable on x=0 when m>1
so its continuously differentiable on m>1

thats the correct solution?

10. Originally Posted by transgalactic
by solving A and B
it is continues on x=0 when m>0
it is differentiable on x=0 when m>1
so its continuously differentiable on m>1

thats the correct solution?
No! A function is continuously differentiable at x = 0 means the derivative at 0 exists and the derivative is continuous at 0 ! I never proved any result about the derivative!

Ok I will do the last one too... But I will let you prove some statements on your own.

Since we have already proved that m > 1 iff f(x) is differentiable at x = 0, for continuous differentiability we need m > 1.
$f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)$

Now first prove that if m > 2, then the function f'(x) is continuous at 0, using (*) and making a similar rule for $\lim_{x \to 0} x^m \cos \frac1{x}$(prove this too!).

Now the only remaining case is m=2. For this case, $f'(x) = \begin{cases} 2x\sin \frac1{x} - \cos \frac1{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$.

Choose the sequence $x_n = \frac1{n \pi}$, and observe that $x_n \to 0$ but $f(x_n) = -(-1)^n$. Thus f(x) is not continuously differentiable at x = 0.

So at the end, let me list the exercises you have to do, to complete the proof rigorously:

(*) exists and equal to 0 iff $m > 0$

(**) $f'(x) = x^{m-2} \left(mx \sin \frac1{x} - \cos \frac1{x}\right)$

(***) $\lim_{x \to 0} x^m \cos \frac1{x}$ exists and equal to 0 iff $m > 0$

(****) Using (**) and (***), prove that if m > 2, then $f'(x)$ is continuous at $x = 0$

11. (*) exists and equal to 0 iff $m > 0$

the only way to prove it
is by cheching the cases of m.
if m<0 then as x goes to zero and m is negative
we get 0.
so the whole limit would be
0*bouded which is bound
so i think i found a mistake in your theory
its working for m differs 0

12. 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
??

13. Originally Posted by transgalactic
(*) exists and equal to 0 iff $m > 0$

the only way to prove it
is by cheching the cases of m.
if m<0 then as x goes to zero and m is negative
we get 0.
so the whole limit would be
0*bouded which is bound
so i think i found a mistake in your theory
its working for m differs 0
Read the question again. It explicitly "shouts" $m \geq 0$.

14. Originally Posted by 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
??
can you explain this thing
??

15. Originally Posted by transgalactic
can you explain this thing
??
That solution you got is of no use to part A. To demonstrate discontinuity at $x_0$, we have to find a sequence $x_n \to x_0$, but the sequence $f(x_n) \not \to f(x_0)$.

Your sequence $x_n \to 0$, but $f(x_n) = \sin 2n \pi = 0$. So it is of no use.

As a side note, I am really sorry transgalactic, that I am failing in explaining this concept to you. I have already given you the answer for this in my previous posts. The sequence you can use is $x_n = \frac2{(2n+1)\pi}$

I am telling the following based on our discussion.....If you take it in the right spirit, I suggest you take tutions. Sometimes a live tutor teaching a new concept is an invaluable experience.

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