1. ## differentiability of Sin(1/x)...

consider: f: R $\rightarrow$ R defined as:
f(x) = $sin(\frac{1}{x})$ if $x \neq 0,$
f(0) = 0.
prove f is not differentiable at 0.

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section A

going by the definition of differentiability, i understand it has to be shown that

$\lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{\sin \frac{1}{x} - sin(0)}{x - 0}$ = $\lim_{x\to 0} \frac{\sin \frac{1}{x}}{x}$ .

DOES NOT EXIST

• i attempted computing the left and right hand limits but i dont think that is the right method?
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section B

i also know if a function is differentiable at 0 then it has to be continuous at 0, which sin(1/x) is not. but i don't have a good idea on how to prove that sin(1/x) is not continuous at 0
??

also, would it be sufficient to just show that sin(1/x) is discontinuous at 0 to prove it is not differentiable at 0?

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please provide assistance greatly appreciated. once again, i'm assuming this must be simple and straight forward for an advanced mathematician

2. Originally Posted by heyo12
also, would it be sufficient to just show that sin(1/x) is discontinuous at 0 to prove it is not differentiable at 0?
Yes.

To show that its not continuous at 0 we need only observe that there are points arbitarily close to 0 where $\sin(1/x)=1$, for instance consoder the points:

$x=\frac{1}{(2n+1/2)\pi}\ \ n=1, 2, 3, ..$

RonL

3. Here is what CaptainBlank did more formally.

Assume that $\sin \frac{1}{x}$ is continous then for any sequence $(x_n)\in \mathbb{R} - \{ 0 \}$ so that $\lim x_n=0$ we must have that $\lim f(x_n) = f(0)=0$. But using the sequence that was shown above we see that fails.

4. Originally Posted by ThePerfectHacker
Here is what CaptainBlank did more formally.

Assume that $\sin \frac{1}{x}$ is continous then for any sequence $(x_n)\in \mathbb{R} - \{ 0 \}$ so that $\lim x_n=0$ we must have that $\lim f(x_n) = f(0)=0$. But using the sequence that was shown above we see that fails.
thnx captain and perfecthacker.

ok this is what i understood from what you said perfecthacker. here's my proof from both your help..

you used the sequential definition of continuity? which implies that:
continuity at 0 = limit at 0 exists and equals f(0)?

PROOF
sin(1/x) is continuous at $x_0 = 0$ if for any sequence $x_n$ $\epsilon$ R, such that $\lim_{n \to \infty}(x_n) = x_0$, we have $\lim_{n \to \infty}f(x_n) = f(x_0)=0$

now let $x_n = \frac{1}{(2n)\pi}$ and $b_n = \frac{1}{(2n + \frac{1}{2})\pi}$

so $\lim_{n \to \infty}(x_n) = \lim_{n \to \infty}(b_n) = 0$

if we consider, $\lim_{n \to \infty}f(x_n)$ we have:
$\lim_{n \to \infty}f(x_n)$ = $\lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n)\pi}}$ = $\lim_{n \to \infty}sin(2\pi n)$ = 0

but, if we consider, $\lim_{n \to \infty}f(b_n)$ we have:
$\lim_{n \to \infty}f(b_n)$ = $\lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}}$ = $\lim_{n \to \infty}sin((2n + 1/2)\pi)$ = 1

we see there are points arbitrarily close to 0 where sin(1/x) = 1.
so this means the limit at 0 doesn't exist
as the limit doesnt exist. the function is not continuous. this implies that it is not differentiable at 0. Q.E.D

is this completely correct? and is there anything else i need to add onto this proof? thanks

5. Originally Posted by heyo12
thnx captain and perfecthacker.

ok this is what i understood from what you said perfecthacker. here's my proof from both your help..

you used the sequential definition of continuity? which implies that:
continuity at 0 = limit at 0 exists and equals f(0)?

PROOF
sin(1/x) is continuous at $x_0 = 0$ if for any sequence $x_n$ $\epsilon$ R, such that $\lim_{n \to \infty}(x_n) = x_0$, we have $\lim_{n \to \infty}f(x_n) = f(x_0)=0$

now let $x_n = \frac{1}{(2n)\pi}$ and $b_n = \frac{1}{(2n + \frac{1}{2})\pi}$

so $\lim_{n \to \infty}(x_n) = \lim_{n \to \infty}(b_n) = 0$

if we consider, $\lim_{n \to \infty}f(x_n)$ we have:
$\lim_{n \to \infty}f(x_n)$ = $\lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n)\pi}}$ = $\lim_{n \to \infty}sin(2\pi n)$ = 0

but, if we consider, $\lim_{n \to \infty}f(b_n)$ we have:
$\lim_{n \to \infty}f(b_n)$ = $\lim_{n \to \infty}sin \frac{1}{\frac{1}{(2n + 1/2)\pi}}$ = $\lim_{n \to \infty}sin((2n + 1/2)\pi)$ = 1

we see there are points arbitrarily close to 0 where sin(1/x) = 1.
so this means the limit at 0 doesn't exist
as the limit doesnt exist. the function is not continuous. this implies that it is not differentiable at 0. Q.E.D

is this completely correct? and is there anything else i need to add onto this proof? thanks
That looks pretty good to me. Though on second thoughts the sequence $x_n$ is probably redundant as we have $f(0)=0$, so we need only display a sequence who's image under $f$ does not go to $0$ to prove that $f(x)$ is discontinuous at $x=0$.

RonL

RonL