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If $f(x)=\begin{cases} x^m\sin\dfrac{1}{x}, & x\ne 0 \\ 0, & x=0 \end{cases}$.

Find $m$ if $f(x)$ is continuous and is not differentiable

My attempt is as follows:-

Let's find the condition of continuity

$$\lim_{x\to0^{+}}x^m\sin\dfrac{1}{x}$$

As $x\rightarrow 0^{+}, \dfrac{1}{x}\rightarrow \infty,\sin\dfrac{1}{x} \text { oscillates in } [-1,1]$

$$m>0$$

$$\lim_{x\to0^{-}}x^m\sin\dfrac{1}{x}$$

As we have the negative base

$$m>0 \cap m\notin \left\{\dfrac{p}{q} | p,q \text { are coprime and } q \text { is even }\right\}\tag{1}$$

Let's find the condition of non-differentiability

$\lim_{h\to 0}\dfrac{h^m\sin\dfrac{1}{h}}{h}$ should not exist

$$\lim_{h\to 0^{+}}h^{m-1}\sin\dfrac{1}{h}$$ $$m\le0$$

$$\lim_{h\to 0^{-}}h^{m-1}\sin\dfrac{1}{h}$$ $$m-1\le 0 \cup m-1\in\left\{\dfrac{p}{q} | p,q \text { are coprime and } q \text { is even }\right\}$$

$$m\le 1 \cup m\in\left\{\dfrac{p+q}{q} | p,q \text { are coprime and } q \text { is even }\right\}\tag{2}$$

Taking intersection of equations $(1)$ and $(2)$

$$\left(m\in(0,1] \cap m\notin \left\{\dfrac{p}{q} | p,q \text { are coprime and } q \text { is even }\right\}\right) \cup \left(m>0 \cap m\notin \left\{\dfrac{p}{q} | p,q \text { are coprime and } q \text { is even }\right\} \cap m\in\left\{\dfrac{p+q}{q} | p,q \text { are coprime and } q \text { is even }\right\}\right)$$

But actual answer is simply $m\in(0,1]$

StubbornAtom
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user3290550
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4 Answers4

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I have to admit I cannot really understand what you are trying to do, and how you come with your conditions for $p$ and $q$.

The first condition is that you need $$ \lim_{x\to0} x^m\sin\frac1x=0. $$ Since the sine is bounded, when $m>0$ we have $\left|x^m\sin\frac1x\right|\leq|x|^m$, so the limit is $0$ and the function is continuous. When $m\leq0$ the function oscilates at $0$ (unboundedly if $m<0$ so the limit doesn't exist).

So far: continuity when $m>0$.

For differentiability, we look at the limit at $0$ of $$ \frac {x^m\sin\frac1x}{x}=x^{m-1}\sin\frac1x. $$ If $m>1$, the limit is zero and so $f$ will be differentiable. When $m<1$, let $x_k=\frac2{\pi(2k+1)}$. Then $$ x_k^{m-1}\sin\frac1{x_k}=\left(\frac{\pi(2k+1)}{2}\right)^{1-m}\xrightarrow[k\to\infty]{}\infty $$ since $m<1$ so $1-m>0$. And when $m=1$ the limit is $1$ we can also choose a sequence so that the limit doesn't exist.

Thus $f$ is differentiable precisely when $m>1$. So,

$f$ is continuous but not differentiable when $0<m\leq 1$.

Martin Argerami
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  • see I will explain you in simple terms, suppose we are considering left hand limit, ok? $\lim_{x\to 0^{-}}x^m\sin \dfrac{1}{x}$, so you can see $x$ is just less than $0$, ok?

    Now if m is $\dfrac{1}{2}$, left hand limit will become undefined right? now you got my point. Because square root of negative doesn't exist, right?

     – user3290550 Dec 14 '19 at 18:07
  • Yeah, with very few exceptions, your function is not defined when $x<0$. The reasonable thing when considering arbitrary powers of a real number is to assume that the number is non-negative. – Martin Argerami Dec 14 '19 at 19:26
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I'd do cases: clearly $\;m>0\;$, otherwise you have $\;\frac1{x^n}\sin\frac1x\;$ , with positive $\;n\;$ , and thus the limit doesn't exist as

$$\;\frac1{x^n}\xrightarrow[x\to 0^\pm]{}\pm\infty\;\;\text{and}\;\;\lim_{x\to 0}\sin\frac1x\;\;\text{doesn't exist}$$

Now, for $\;0\le m\le1\;$ , we get that

$$\lim_{x\to0} x^m\sin\frac1x=0\;\;\text{since sine is bounded, but}\;\;f'(0)=\lim_{x\to0}\frac{x^m\sin\frac1x}x=\lim_{x\to0}x^{m-1}\sin\frac1x$$

and this last limit doesn't exist by the reasons above as $\;m-1\le0\;$.

Thus, it must be $\;m>1\;$ , and then

$$f'(0)=\lim_{x\to0}x^{m-1}\sin\frac1x=0\;,\;\;\text{ since}\;\;m-1>0$$

DonAntonio
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  • can you tell me what's wrong in my approach and you have not considered when x is negative and m is like $\dfrac{1}{2}$ – user3290550 Dec 14 '19 at 17:55
  • Why would the sign of $;x;$ be important at all?! All that matters is that $;x\to0;$ and the limit of a function converging to zero times a bounded one is zero. – DonAntonio Dec 14 '19 at 17:56
  • As your approach is way to messy and it seems to deal with rationals in reduced form and other things that I can't understand how are relevant here. – DonAntonio Dec 14 '19 at 17:57
  • if function needs to be continuous, left hand limit and right hand limit both should exist – user3290550 Dec 14 '19 at 17:57
  • @user3290550 Exactly! And that's what the above approach shows in case $;0<m\le1;$ ... – DonAntonio Dec 14 '19 at 17:58
  • again if you see the left hand limit at $0$, we are dealing with x just less than $0$, right?

    now you got my point

     – user3290550 Dec 14 '19 at 17:59
  • @user3290550 Yes...and what?!! It doesn't make any difference! The limit $;x^m\sin\frac1x=0;$ whenever $;m>0;$ no matter whether $;x<0;$ or not...I don't really understand your insistence in this... – DonAntonio Dec 14 '19 at 18:01
  • see if $x$ is just less than zero, ok? it means $x$ is negative, right? now m is – user3290550 Dec 14 '19 at 18:02
  • BTW, what does "$;x\in0;$" mean? Is it "different from zero" or $;x>0;$ ? – DonAntonio Dec 14 '19 at 18:02
  • see I will explain you in simple terms, suppose we are considering left hand limit, ok? $\lim_{x\to 0^{-}}x^m\sin \dfrac{1}{x}$, so you can see $x$ is just less than $0$, ok?

    Now if m is $\dfrac{1}{2}$, left hand limit will become undefined right? now you got my point. Because square root of negative doesn't exist, right?

     – user3290550 Dec 14 '19 at 18:05
  • @user3290550 Indeed so, and thus it is way easier: your function is undefined on any left neighborhood of zero ad thus it must be defined on the positive real numbers, and thus the question must be only about right continuity and right differentiability. – DonAntonio Dec 14 '19 at 18:13
  • let me make some cases in general:- Case 1: only one of right hand limit or left hand limit is defined and the other one is undefined, does it means limit exist? Case 2: none of right hand limit or left hand limit is defined, can we say limit exists? – user3290550 Dec 14 '19 at 18:22
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The funcion $f$ is differentiable at $0$ if and only if the limit$$\lim_{x\to0}\frac{f(x)-f(0)}x\tag1$$exists. But, since $f(0)=0$, the assertion that the limit $(1)$ exists is equivalent to the assertion that the limit$$\lim_{x\to0}x^{m-1}\sin\left(\frac1x\right)$$exists. And this happens if and only if $m-1>0$.

José Carlos Santos
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  • can you tell me what's wrong in my approach and you have not considered when x is negative and m is like $\dfrac{1}{2}$ – user3290550 Dec 14 '19 at 17:55
  • When $m<0$, it is clear that $f$ isn't even continuous. – José Carlos Santos Dec 14 '19 at 17:57
  • Besides, the behaviour of the limit is the same when $h$ goes to $0$ from above and from below. – José Carlos Santos Dec 14 '19 at 17:58
  • see I will explain you in simple terms, suppose we are considering left hand limit, ok? $\lim_{x\to 0^{-}}x^m\sin \dfrac{1}{x}$, so you can see $x$ is just less than $0$, ok?

    Now if m is $\dfrac{1}{2}$, left hand limit will become undefined right? now you got my point. Because square root of negative doesn't exist, right?

     – user3290550 Dec 14 '19 at 18:08
  • Right. Then $f$ is actually undefined when $x<0$. – José Carlos Santos Dec 14 '19 at 18:10
  • so $\lim_{x\to 0}x^m\sin \dfrac{1}{x}$ will not exist right? for such values of $m$ $\dfrac{1}{2},\dfrac{3}{2}\cdots\cdots$ as left hand limit is getting undefined – user3290550 Dec 14 '19 at 18:13
  • @user3290550 It is moot to ask about that limit as the function doesn't exist when $;x<0;$ unless $;m=0;$...and thus it isn't even continuous. The question thus must be about right continuity and differentiability at zero. – DonAntonio Dec 14 '19 at 18:14
  • Yes, the limit exists. For instance, the limit $\lim_{x\to0}\sqrt x$ exists; it is equal to $0$. – José Carlos Santos Dec 14 '19 at 18:14
  • @JoséCarlosSantos That's been discussed several times in the past, and it all depends on definitions. You seem to take $;\lim\limits_{x\to0}\sqrt x;$ as the limit within the definition domain of the function...and that's fine. but not all oblige by the same and things must be made clear. – DonAntonio Dec 14 '19 at 18:18
  • let me make some cases in general:-

    Case $1$: only one of right hand limit or left hand limit is defined and the other one is undefined, does it means limit exist?

    Case $2$: none of right hand limit or left hand limit is defined, can we say limit exists?

     – user3290550 Dec 14 '19 at 18:18
  • @DonAntonio I have never been exposed to another possibility. – José Carlos Santos Dec 14 '19 at 18:19
  • @JoséCarlosSantos I have and it is not a universal thing. Read some comments and answer here, for example: https://math.stackexchange.com/questions/637280/limit-of-sqrt-x-as-x-approaches-0 – DonAntonio Dec 14 '19 at 18:49
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Your approach in correct in all aspects except your considering fractional powers with even denominators as base. The original question does not imply any condition to the domain of the function, it asks you to define a function. The domain is defined after you define the function. Here, you need not necessarily take the domain as $\Bbb R$.
In those functions with $m\in \left\{\dfrac{p}{q} | p,q \text { are coprime and } q \text { is even }\right\}$, the domain would be simply defined as $\Bbb R^{+}\cup \{0\}$ , and in that case continuity and differentiability would be checked from the right side. See:- for limit at end point of domain. And those functions will not pose any problem to the conditions given question.

Harshil
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