Proving the limit of $\sin(\frac{1}{x})$ doesn't exist as $x$ tends to $0$

Question

I want to do this by contradiction from the definition.

Assume for sake of a contradiction that the limit, $l$, exists, where $l$ is an element of the extended real line. Let $0$ be an accumulation point of of the domain. Notice that |$\sin\left(\frac{1}{x}\right)$| $\leq$ $1$ for all $x$ real numbers apart from $0$, so $l$ can't be ±$\infty$.

Let $\epsilon$ = $\frac{1}{2}$. If $0$ $<$ $x$ < $\delta$, when $ 0+2k\pi \leq x \leq \pi + 2k\pi$ then |$1-l$| < $\frac{1}{2}$. If $-\delta$ < $x$ $< 0$, when $x$ < $0$, $ \frac{-\pi}{2} + 2k\pi \leq x < 0 + 2k\pi$, then |$-1-l$| < $\frac{1}{2}$.

We now have, $2$ = |$1-l+1+l$| $\leq$ |$1-l$| + |$1+l$| $=$ |$1-l$| + |$-1-l$| $<$ $\frac{1}{2}$ $+$ $\frac{1}{2}$ = $1$, which is a contradiction.

Therefore, the limit doesn't exist.

What mistakes have I made. I was considering using change of variable $y$ = $\frac{1}{x}$, so the question would be the limit of $y$ approaching $\infty$ of $\sin(y)$.

Made some edits above. Is it possible for me to prove this using this method with some minor changes ?

Answer 1

If the limit existed, we would have that $\lim_{n\to \infty} \sin \frac{1}{a_n}$ would be the same for any sequence $a_n \to 0$. However, $$ \lim \sin\frac{1}{\frac{1}{\pi n}} = \lim \sin (n \pi) = 0 $$

$$ \lim \sin\frac{1}{\frac{1}{\frac{\pi}{2} +2\pi n}} = \lim \sin (\frac{\pi}{2} + 2 n \pi) = 1. $$

This means that in any neighbourhood of $x=0$ you can find values of $\sin \frac 1x$ as close as you wish to 0, and also to 1, which renders the definition of limit impossible to satisfy.

Answer 2

If you want to use the definition of limits: For all $l \in \mathbb{R}\setminus\{1\}$, there exists $\epsilon_0=|1-l|/2$, $\forall \delta > 0$, there exists $x_0= (\frac{\pi}{2}+k2\pi)^{-1} \in (0, \delta)$ for $|k| \in \mathbb{N}$ large enough, but $|\sin(\frac{1}{x_0})-l|= |\sin (\frac{\pi}{2}+k2\pi)-l| = |1-l| > \epsilon_0$.

For $l=1$, there exists $\epsilon_0=1/2$, $\forall \delta > 0$, there exists $x_0= (k2\pi)^{-1} \in (0, \delta)$ for $|k| \in \mathbb{N}$ large enough, but $|\sin(\frac{1}{x_0})-l|= |\sin (k2\pi)-l| = 1 > \epsilon_0$.