Prove $\lim_{x \to 0}\frac{x}{x-\lfloor \sin x \rfloor}$ doesn't exist.
I can prove the same claim using Heine's definition by showing that sequences $\{x_n\}=\frac{\pi}{2n}$ and $\{x_m\} = -\frac{\pi}{2m}$ converge to 0, but the values of function at these sequences diverges.
In the proof from definition I can get about this far:
Suppose the limit exists at $x_0 = 0$, then choose $\epsilon = \frac{1}{2}$ (I know this should fail because as $x$ approaches zero from left, it tends to zero and as $x$ approaches zero from right it tends to 1). Since $x \in (-\sigma, \sigma)$ this should hold:
$$
-\frac{1}{2} < \frac{\sigma}{\sigma – \lfloor \sin \sigma \rfloor} – L < \frac{1}{2} \\
-\frac{1}{2} < \frac{-\sigma}{\lfloor -\sin \sigma \rfloor -\sigma} -L < \frac{1}{2}
$$
Here my problem starts. I believe that from the first equation I'd need to derive that $L \geq 1$ and from the second: $L \leq 0$. But there's no way I can get at these values.
More importantly, I'm trying to discover a general procedure for solving such questions, here's what I "discovered" so far:
- Guess $\epsilon$ (by calculating several values of the function near $x_0$).
- Assume $x \in (-\sigma, \sigma)$.
- Write two inequalities: $-\epsilon < f(\sigma) – L < \epsilon$ and $-\epsilon < f(-\sigma) – L < \epsilon$.
- Apply series of algebraic transformation to both inequalities until contradiction is apparent.
- The proof is complete at this point.
Does this sound like a reasonable approach?
Best Answer
Hint: when $0 < x < \pi$, $\lfloor \sin x \rfloor = 0$, so that $\frac{x}{x - \lfloor \sin x \rfloor} = 1$. When $-\pi < x < 0$, $\lfloor \sin x \rfloor = -1$, so that $\frac{x}{x - \lfloor \sin x \rfloor} = \frac{x}{x + 1}$. Take $\epsilon = \frac{1}{2}$ (say) and see if you can find a $\delta >0$ such that when $-\delta < x < 0$, then $\left|\frac{x}{x+1}\right| < \epsilon$.
See the comments below for more on the methodological question about general procedures for solving this kind of problem.