Proof verification $\lim_{x \rightarrow 0}{\sin(\frac{\pi}{x})}$ does not exist.

Question

Prove that $\lim_{x \rightarrow 0}{\sin(\frac{\pi}{x})}$ does not exist. Usually textbooks prove this by using two sequences that converge to different limits. I'd like to know if the following reasoning is valid.

Proof: Notice that $\lim_{x \rightarrow 0}{\sin(\frac{\pi}{x})}$ = $\lim_{x \rightarrow 0}{\sin(\frac{\pi}{x})} \cdot \frac{\frac{\pi}{x}}{\frac{\pi}{x}} = \lim_{x \rightarrow 0}{\frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}} \cdot \frac{\pi}{x}$.

From the fundametal trigonometric limit, we know that $\lim_{x \rightarrow 0}{\frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}} = 1$.

Also, $\lim_{x \rightarrow 0^+}{\frac{\pi}{x}} = +\infty$ and $\lim_{x \rightarrow 0^-}{\frac{\pi}{x}} = -\infty$.

Hence, $\lim_{x \rightarrow 0^+}{\frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}} \cdot \frac{\pi}{x} = +\infty$ and $\lim_{x \rightarrow 0^-}{\frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}} \cdot \frac{\pi}{x} = -\infty$.

Therefore, $\lim_{x \rightarrow 0}{\frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}} \cdot \frac{\pi}{x} = \lim_{x \rightarrow 0}{\sin(\frac{\pi}{x})}$ does not exist.

Answer 1

No, this is not correct. Since $x\mapsto\sin\left(\frac\pi x\right)$ is a bounded function, there is no way you can have $\lim_{x\to0^\pm}=\pm\infty$.

If $\lim_{x\to0}f(x)=0$, then, yes, $\lim_{x\to0}\frac{\sin(f(x))}{f(x)}=1$. But you don't have $\lim_{x\to0}\frac\pi x=0$.

Answer 2

$\lim_{ x \to 0}\frac{sin(\frac{\pi}{x})}{\frac{\pi}{x}} \neq 1$.

Look here for a explanation. In the way you have done, the limit for both cases $x \to 0^{+}$ and $x \to 0^{-}$ will be zero.

Answer 3

$$L=\lim_{x\to 0}~ \sin(\pi/x)$$ is a real, bounded in $[-1,1]$ but uncertain number. So the limit does not exist.