# continuety parameter question..

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• January 18th 2009, 06:50 AM
transgalactic
continuety parameter question..
i tried to solve it
but i didnt get any parameter value
??
http://img403.imageshack.us/img403/5...9192yz6.th.gif
• January 18th 2009, 07:08 AM
Isomorphism
http://img403.imageshack.us/img403/5395/25739192yz6.gif

Here's a thought: what happens when m=0 ?
• January 18th 2009, 07:20 AM
transgalactic
0^0 *bounded is an undefined form
• January 18th 2009, 07:46 AM
Isomorphism
Quote:

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)
• January 18th 2009, 09:29 AM
transgalactic
ok so its 1*bounded wich is not defined either

??
• January 18th 2009, 09:34 AM
Isomorphism
Quote:

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 $f(x_n) = (-1)^n$
• January 18th 2009, 12:33 PM
transgalactic
how does it finds the values of the m parameter
• January 18th 2009, 05:26 PM
Isomorphism
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 :)
• January 18th 2009, 11:28 PM
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?
• January 18th 2009, 11:59 PM
Isomorphism
Quote:

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:

(*) http://www.mathhelpforum.com/math-he...9d641f57-1.gif 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 $x = 0$

• January 19th 2009, 02:30 AM
transgalactic
(*) http://www.mathhelpforum.com/math-he...9d641f57-1.gif 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
• January 19th 2009, 03:04 AM
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
??
• January 19th 2009, 04:45 AM
Isomorphism
Quote:

Originally Posted by transgalactic
(*) http://www.mathhelpforum.com/math-he...9d641f57-1.gif 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" $m \geq 0$.
• January 19th 2009, 05:16 AM
transgalactic
Quote:

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
??
• January 19th 2009, 06:42 AM
Isomorphism
Quote:

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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