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Let $f:[0, \pi] \rightarrow \mathbb{R}$ be defined by $f(0)=0$ and $f(x)=x \sin \frac{1}{x}-\frac{1}{x} \cos \frac{1}{x}$ for $x \neq 0$. Is $f$ continuous ?

my method was that x $\sin \frac{1}{x}$ has limit value as zero as x tends to zero . So we can focus on just the $\frac{1}{x} \cos \frac{1}{x}$ , that term is having the form if infinity * some number between -1 and 1 , both inclusive , so we can say its indeterminate hence limit doesnt exist ? Or we cannot do this method as when we apply partwise limit check we need to ensure both limit exists?

ProblemDestroyer
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    I recommend that you try to find a sequence of numbers $x_n\to0$ with the property that $\cos(1/x_n)=1$ for each $n$. What then are the resulting values of $f(x_n)$? – Mark McClure May 05 '22 at 12:44
  • Okay i will surely try it , from a glance it seems we need to take x_n to be 1/(2pi n), but can you once tell if my original method was correct or not ? I mean i think its wrong becaue i took limit f(x) = limit of both indicidual limits without showing that both should exist then only we could have had applied this method ? @MarkMcClure – ProblemDestroyer May 05 '22 at 12:51
  • It looks like there's an answer that correctly addresses the problematic issues with your approach. I would also point out that $\sin(x)=0$ when $\cos(x)=1$. Thus, there's no need to break the problem up by dealing with the summands separately. – Mark McClure May 05 '22 at 13:49
  • I am sorry but yeah i agree that sinx = 0 when cosx =1 , but how it tells that summand is not needed to be considered separately we dont know the exact varation as its of form cos(infinity) and sin(infinity) Sir ? – ProblemDestroyer May 05 '22 at 13:59
  • Your method is correct (+1), but you need to fix it a litte. The last term consists of product of two factors one of which tends to $\infty$ and the other one oscillates between $-1$ and $1$ (you used $0,1$ inclusive and that is insufficient to conclude). So effectively the term oscillates infinitely. – Paramanand Singh May 05 '22 at 19:04
  • Thanks for confirming @Paramanand Singh – ProblemDestroyer May 05 '22 at 19:05

2 Answers2

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It is a mistake to say the function 'is indeterminate' – the expression you're considering is defined for $x\ne 0$, where $\frac 1x$ is well defined, its cosine $\cos\frac 1x$ is well defined and their product is well defined, too.

You should rather point out that cosine's range $[-1,1]$ (not $[0,1],$ btw) allows values of the subexpression to reach from $-\frac 1{|x|}$ to $\frac 1{|x|}$, which gets unbounded when $x$ approaches zero.

That implies $f$ is unbounded around $x=0$ which breaks the condition in the limit's definition.
As a result, $f$ has no limit at $x=0.$

CiaPan
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  • Is it always true btw Sir that suppose h(x) = f(x) +g(x) and we are assuming limit of h(x) to exist , we also know that limit of f(x) surely exists and i have the value of that already . , If we showed that limit g(x) doesnt exist. This will imply limit h(x) which we assumed to exist is defintely not existing? – ProblemDestroyer May 05 '22 at 13:34
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    @ProblemDestroyer: I think that you should state the definition of 'discontinuous' in the OP. Some people use '$f$ is discontinuous at $a$' to mean that $f$ is either not defined at $a$ or $f$ is not continuous at $a$ (if it is defined there). User CiaPan seems to be using the definition that -$f$ is said to be discontinuous at $a$ if $a$ is in domain of $f$ and $f$ is not continuous at $a$. – Koro May 05 '22 at 13:41
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    @ProblemDestroyer: Please note that with the former definition $x\mapsto 1/x$ is discontinuous at $0$ but w.r.t. the other definition $x\mapsto 1/x$ is neither continuous nor discontinuous at $0$. – Koro May 05 '22 at 13:46
  • @Koro you are very right , i should have mentioned the def of discontinuous i am using . But yeah i agree with all yours points , very very thanks for providing insights – ProblemDestroyer May 05 '22 at 13:56
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    @ProblemDestroyer (1/2) User Koro is right. I am so used to the "limit=value" definition that I overlooked the other possibility. They are both mentioned e.g. in Wikipedia article Continuous function in section Definition in terms of limits of functions and in Definition in terms of neighborhoods, respectively. – CiaPan May 05 '22 at 14:00
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    @ProblemDestroyer (2/2) The very same article uses them both inconsitently, for example the caption of an image in Real functions, Definition says $x\mapsto 1/x$ is continuous over its domain, but discontinuous at $x=0$, whilst a similar rational function in Construction of continuous functions is said to be continuous, just not defined at $x=-2.$ – CiaPan May 05 '22 at 14:00
  • Understood @CiaPan , can you once check what i posted in the first comment in this answer of yours Sir? – ProblemDestroyer May 05 '22 at 14:13
  • @ProblemDestroyer No, if functions $f, g, h$ are defined on the same neighborhood of some value $p,$ and they satisfy $h(x)=f(x)+g(x),$ and both $f$ and $g$ have a limit at $p$, then their sum also has a limit there and the limit of the sum equals the sum of limits: $\lim_{x\to p}h(x) = \lim_{x\to p}f(x) + \lim_{x\to p}g(x).$ So it's not possible for one of them to have no limit while other two have limits. OTOH it's possible one has a limit and two do not have limits. – CiaPan May 05 '22 at 15:44
  • Yeah Sir so what i am saying was that by assumption of h(x) limit being exist , although f(x) is existing as earlier said, we get limit of g(x) to be existing from the sum and difference rule of limits , but if that g(x) limit is not existing , so that is contradicting the first assumption , hence limit of h(x) is not existing ?? I am appying this here where we know f(x)= xsin(1/x) has finite limit , while h(x) we at start assumed it , but if that is true , g(x) = 1/x cos(1/x) should also have a finite limit but thats not true , hence h(x) limit doesnt exist ? – ProblemDestroyer May 05 '22 at 15:51
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In order for $$ f(x) = \begin{cases} x \sin \left(\frac{1}{x}\right)-\frac{1}{x} \cos \left( \frac{1}{x}\right) & x\neq0 \\ 0 & x=0 \end{cases} $$ to be continuous at $x=0$, we must have $$ \lim_{x\to0}f(x) = 0. $$ In particular, we must have $$ \lim_{n\to\infty}f(x_n) = 0, $$ for every real sequence $(x_n)$ that tends to zero.

Now, consider $\displaystyle x_n = \frac{1}{2n\pi}$. Note that $$ f\left(\frac{1}{2n\pi}\right) = \frac{1}{2n\pi}\,\sin(2n\pi) + 2n\pi\cos(2npi) = 2n\pi \not\to 0. $$ Thus, for this particular sequence satisfying $x_n\to0$, we don't have $f(x_n)\to0$. Thus, $f$ is not continuous at zero.

Mark McClure
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