1. If you tell me clearly, what part of my proof in post #8 and post #10, you have not understood, I will try to clarify it. Dont simply say Undefined * bounded etc

Learn from your notes that testing for continuity is not just substituting the value at that point.

2. Originally Posted by Isomorphism
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.
i cant see from where you came up with that sequence??
$x_n = \frac2{(2n+1)\pi}$

3. Originally Posted by transgalactic
i cant see from where you came up with that sequence??
$x_n = \frac2{(2n+1)\pi}$
Well I wanted $x_n$ to converge to 0 and $f(x_n)$ to not go to zero. I realised that trigonometric function alternates between +1 and -1, and for sin it happens when the angle is $\frac{(2n+1)\pi}{2}$. Thus I used it to my advantage. So by defining that sequence $f(x_n) = (-1)^n$, which does not have a limit.

4. ok i understood your last post
regarding m=0 case
i will try to prove what you asked

5. Hi,

I shall try to summarize the answers to your questions. You should also read Isomorphism's posts again from the beginning.

For a given $m\in\mathbb{N}$, we consider the function $f:x\mapsto \left\{\begin{array}{cc}x^m\sin\frac{1}{x} & \mbox{if }x\neq 0\\ 0 & \mbox{if }x=0\end{array}\right.$ defined on $\mathbb{R}$.

First remark is that $f$, when restricted to the intervals $(-\infty,0)$ or $(0,+\infty)$, is infinitely differentiable (i.e. it can be differentiated any number of times). It is indeed obtained by product and composition of functions that are infinitely differentiable on their respective domain. And we can compute $f'(x)$ for $x\neq 0$ using usual rules.

It is not clear however what happens at 0 since $\frac{1}{x}$ diverges when $x\to0$.

a) The continuity at $0$ means that $f(x)\to f(0)$ when $x\to 0$.
- Like you said, if $m>0$, then $f$ is the product of a bounded function (the sine) by a function converging to 0 at 0, so that $f$ converges to 0 at 0. Explicitly, we have $|f(x)-f(0)|=|f(x)|\leq |x|^m$ and the right-hand side converges to 0 when $x$ tends to 0. Thus $f$ is continuous.
- When $m=0$, we have $f(x)=\sin\frac{1}{x}$ for $x\neq 0$. Plot this function on your calculator. When $x$ goes down from 1 to 0, $\frac{1}{x}$ goes up (continuously) from 1 to $+\infty$, so that $\sin\frac{1}{x}$ oscillates infinitely many times (it is like the whole sine curve on $[1,+\infty)$ were "compressed" (backwards) into the interval $(0,1]$, with increasing compression rate on the neighbourhood of 0). Once you understand what the graph looks like, it is pretty obvious that $f$ won't have a limit at 0. One reason is that $f$ achieves the value 1 at points arbitrarily close to 0, whereas the continuity would mean that $f$ is close to 0 on small neighboorhoods of 0. The rigorous proof goes through a sequence like the one introduced by Isomorphism: the points $x$ such that $f(x)=1$ are such that $\frac{1}{x}=\frac{\pi}{2}+2k\pi$ for some integer $k$. Thus, if $x_k=\frac{1}{\frac{\pi}{2}+2k\pi}$, we have $f(x_k)=1$ for all $k$, while obviously $x_k\to_{k\to\infty} 0$. If $f$ was continuous at 0, we would have $\lim_k f(x_k)=f(0)=0$, whereas we have $\lim_k f(x_k)=1$: contradiction. Hence $f$ is not continuous at 0.

b) Differentiation at 0. Remember a function needs to be continuous in order to be differentiable. As a consequence, we only have to deal with the case $m>0$.
$f$ is differentiable at 0 iff the ratio $\frac{f(x)-f(0)}{x-0}$ has a limit when $x\to 0$, $x\neq 0$. Note that this ratio equals $\frac{f(x)}{x}=x^{m-1}\sin\frac{1}{x}$, and $m-1\in\mathbb{N}$. Again, we know from the previous part that, at 0, this function converges to 0 if $m-1>0$ and it diverges if $m-1=0$. So that $f$ is differentiable if and only if $m>1$, i.e. $m\geq 2$ since $m$ is an integer. In addition, in this case, we have $f'(0)=0$ (the value of the limit).

c) Continuity of the derivative at 0. The function needs to be differentiable, so we restrict to $m\geq 2$. First compute the derivative of $x\mapsto x^m \sin\frac{1}{x}$ on $\mathbb{R}\setminus\{0\}$, like you usually do (Isomorphism did it). Then look at its limit at 0. The derivative is continuous iff this limit exists and is 0 (since we said that $f'(0)=0$). The derivative is found to be the sum of two terms that are pretty similar to $f$. One of them converges to 0 for any $m\geq 2$, while the other one converges to 0 iff $m>2$. Hence the sum converges to 0 iff $m>2$, i.e. $m\geq 3$.

I hope this clarifies things in your mind...

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