Proving $\lim_{x\to0}\sin\left(\frac{\pi}{x}\right)$ is undefined using the $\epsilon$-$\delta$ definition of a limit

Question

It is well known that $$\lim_{x\to0}\sin\left(\frac{\pi}{x}\right)$$ is undefined, which is intuitively true since the function is periodic and oscillates between $1$ and $-1$ as $x$ approaches zero, and never stays close to any particular value.

However, is there a way to rigorously prove this using the $\epsilon$-$\delta$ definition of a limit?

I have recorded my (slightly roundabout) attempt by contradiction as follows -

Suppose that $\lim_{x \rightarrow 0} \sin\left(\dfrac{\pi}{x}\right)=L$ for some finite value of $L$, and let $x_n=\dfrac{2}{2n+1}, n \in \mathbb{Z_0^+}$. Since $\lim_{n \rightarrow \infty} x_n=0$, it follows that $\lim_{n \rightarrow \infty} \sin\left(\dfrac{\pi}{x_n}\right)=L$ too. This implies that $\forall \ \epsilon >0, \exists \ M \in \mathbb{Z_0^+}$ such that $\forall \ n>M, \left \lvert \sin\left(\dfrac{\pi}{x_n}\right) -L \right \rvert < \epsilon.$

Now, pick some arbitrary $M$. Suppose for $n=M+1, \sin\left(\dfrac{\pi}{x_n}\right)=1$, and for $n=M+2, \sin\left(\dfrac{\pi}{x_n}\right)=-1$. The former implies $\lvert 1-L \rvert < \epsilon \ (\forall \ \epsilon >0)$, which means that $L$ has to be equal to $1$. On the other hand, the latter implies that $\lvert -1-L \rvert < \epsilon \ (\forall \ \epsilon >0)$, which means $L$ has to be equal to $-1$. But clearly, $L$ cannot be both equal to $1$ and $-1$ at the same time, and since $M$ was chosen arbitrarily, we have a contradiction. Note that a similar argument holds if we had chosen $\sin\left(\dfrac{\pi}{x_n}\right)=-1$ when $n=M+1 $ instead.

Is my argument valid, or is more justification needed?

Answer 1

$\lim_{x\rightarrow 0 } \sin\left(\dfrac{\pi}{x}\right)$

Think of two sequences $x_n=\frac{1}{(2n+1/2)}$ and $x'_n=\frac{1}{(n+1)}$ such that when $n \rightarrow \infty$ bot $x_n,x'_n$ tend to 0. As $f(x)=\sin(\pi/x)$, then $f(x_n)=1$ and $f(x'_n)=0$ being unequal, the said limit does not exist.

Answer 2

The image of the sine function is the closed interval $[0,1]$, so for a given value $L\in [0,1]$ there is a $b\in\mathbb{R}$ such that $\sin(b)=L$. We may assume $b>0$. By periodicity of the sine function $\sin(b+2\pi n)=L$ for all $n\in\mathbb{Z}$. We may set $x_n=\frac{\pi}{b+2\pi n}$ for $n\in\mathbb{N}$ which is well-defined, since the denominator is strictly positive. The sequence converges to $0$ and satisfies $\frac{\pi}{x_n}=b+2\pi n$ and therefore $\sin\left(\frac{\pi}{x_n}\right)=\sin(b+2\pi n)=\sin(b)=L$.