Prove $\sin(1/x)$ is discontinuous at 0 using epsilon delta definition of continuity

Question

Let $$f(x) = \begin{cases} 0 &\text{ if $x=0$,}\\ \sin(1/x) &\text{ otherwise.} \end{cases} $$ Prove that $f$ is discontinuous at $0$ using the $\epsilon \delta$ definition of continuity.

I know that the $\epsilon, \delta$ criterion is as follows: for all $\epsilon>0$ there exists $\delta \gt 0$ such that for all $x\in A$, if $|x-x_0|\lt \delta$ implies that $|f(x)-f(x_0)|\lt \epsilon$.

I am unsure how to go about writing the proof and any help would be appreciated! Thank you.

Answer 1

Take the sequence $x_n = \frac{2}{(4n+1) \pi}$. Then $x_n \to 0$ so if $f$ was continuous, we would have $f(x_n) \to f(0) = 0$. However, $f(x_n) = 1$ for all $n$, so clearly we can't have $f(x_n) \to 0$ and thus $f$ isn't continuous.

Using $\epsilon, \delta$: set $\epsilon = 1/2$. Then for any $\delta >0$, we can find $n$ so that $x_n = \frac{2}{(4n+1) \pi} < \delta$. Then $\lvert f(x_n) - f(0) \rvert = \lvert f(x_n) \rvert = 1 \ge \epsilon.$ That is, for this specific $\epsilon$, there is no $\delta > 0$ which guarantees $\lvert f(x) - f(0) \rvert < \epsilon$ for all $\lvert x - 0 \rvert < \delta$. Hence $f$ is not continuous at $0$.

Answer 2

Let y=1/x . The limit as y goes to infinity of sin(y) is undefined because sin is periodic. But this is the limit from the right of sin(1/x) as x goes to 0 (and is still undefined). In particular the right-hand limit is not 0 and thus the general limit is not 0. It follows that the function is discontinuous at 0.

Answer 3

The key idea of the $\epsilon-\delta$ or the formal continuity definition is that if for an arbitrarily small $\epsilon$ neighbourhood "of" $f$ around $f_0$ there exists at least some $\delta$ neighbourhood "in" $x$ around $x_0$ that always maps to within that $f$ space neighbourhood, $f$ is continuous.

So to show a function is discontinuous, one approach is to show that no matter the $\delta$ disk one tries to choose, the image of that disk always has a minimum size. As a result, $\epsilon$ neighbourhoods "of" $f$ that are smaller, are ruled out, breaking continuity.

We have been asked about continuity around $0$ so let's look at a $\delta$ neighbourhood of $0$ and try to estimate the size of its image. One point that this neighbourhood certainly contains is its center, $0$, so the image of this neighbourhood certainly contains $f(0)$ ($=0$ here but I leave it symbolic for arbitrary choices). Now are there any other "easy" points we know the value of $\sin$ at? Of course - integer multiples of $\pi/2$. So, whenever $1/x$ is an integer multiple of $\pi/2$, $\sin(1/x)$ will take on values $0$ or $1$ (or $-1$ but I am ignoring negative values to keep everything simpler. Also, if you do not like these 'integer multiples of $\pi/2$, you can choose any other value of $x$ that you have memorized $\sin$ by heart at, provided $\sin(1/ x)$ isn't $f(0)$ at all of them)

How does this solve the question? If every $\delta$-neighbourhood contains any of those 'easy' points, then the image set of that disk will necessarily contain $\{0,1,f(0)\}$. Since the size of such a disk is at least one as $|1-0|=1$ regardless of $f(0)$, this dooms $f$ to discontinuity. The epsilons who are less than $1$ will never find an appropriate $\delta$ (and forever roam in loneliness)

So, does our neighbourhood contain such points? Yes, since you can always find an $x$ in any $\delta$- neighbourhood around $0$ such that $\sin(1/x)=0 ~ \text{or} ~ 1$. These are precisely the $x<\delta$ that also satisfy $\frac{1}{x}=n \pi/2$ for some $n\in \mathbb{N}$

---

PS: I wrote this verbose, relaxed explanation just to convince myself of and possibly help future beginners about what was going on. In essence, it says nothing more and certainly less, than User8128's terse answer.