I am currently working through Apostol's Calculus, and I was hoping that someone could verify that the proof that I wrote for one of the problems actually proves the assertion.

Prove that $\not\exists A \in \mathbb{R}$ s.t $\lim_{x\to 0} \sin(1/x) = A$.

Proof. Assume that $\exists A\in\mathbb{R}\ $ s.t. $\lim_{x\to0} \sin(1/x) = A.$ There are three cases.

Case 1: $|A| > 1$

By the definition of a limit, $\forall\epsilon>0$, $\exists\delta>0$ s.t.

$$ |\sin(1/x) - A| < \epsilon \text{ whenever }\ |x| < \delta .$$

Restrict $\epsilon$ s.t. $\epsilon < |A| - 1$. Noting that in the interval $(-\delta,\delta)$, $\sin(1/x)$ is at most $1$,

$|\sin(1/x) - A| < |1-A| = |A-1|<\epsilon<|A|-1$.

However, $|A - 1| \geq |A| - 1.$ Thus, $|A| \not> 1$.

Case 2: $|A| = 1$

Once again, by the definition of a limit, $\forall\epsilon>0$, $\exists\delta>0$ s.t.

$$ |\sin(1/x) - A| < \epsilon \text{ whenever }\ |x| < \delta .$$

Restrict $\epsilon$ s.t. $\epsilon < 1$. Noting that at some point in the interval $(-\delta,\delta)$, $\sin(1/x) = -A,$

$|-A - A|=2|A|=2 <\epsilon < 1$,

which is clearly false. Thus, $ |A| \not= 1.$

Case 3: $|A| < 1$

Once again, by the definition of a limit, $\forall\epsilon>0$, $\exists\delta>0$ s.t.

$$ |\sin(1/x) - A| < \epsilon \ \text{ whenever } \ |x| < \delta .$$

Restrict $\epsilon$ s.t. $\epsilon < 1 - |A|$. Noting that $|\sin(1/x)| - |A| \leq |\sin(1/x) - A|$,

$|\sin(1/x)| - |A| \leq |\sin(1/x) - A| < \epsilon < 1 - |A|$.

Thus, $|\sin(1/x)| < 1$. However, in the interval $(-\delta, \delta), \exists x$ s.t. $|\sin(1/x)| = 1$. Thus, $|A| \not< 1$.

As shown through these cases, assuming that $A$ exists always results in a contradiction. Thus, $\lim_{x\to0}\sin(1/x)$ does not exist.

It seems a bit too long. I have seen the other proofs that use sequences; however, Apostol hinted at the use of proof by contradiction. Is there any way I could condense/improve this proof?

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    $\begingroup$ A concise but tangent proof is to consider two distinct sequences $\{x_n\}$ and $\{y_n\}$, both converging to $0$, but result in different limits. Before applying this proof, think deeply what the statement "the limit of $\sin(1/x)$ doesn't exist as $x \to 0$ " means, or more generally, what $f(x)$ doesn't have a limit when $x \to x_0$ means. $\endgroup$ – Zhanxiong Jul 10 '15 at 2:23
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    $\begingroup$ Note that there is no actual reason to split into cases. $\endgroup$ – Amitai Yuval Jul 10 '15 at 2:27
  • $\begingroup$ Do you know Heine's Theorem? $\endgroup$ – Vim Jul 10 '15 at 2:33

Suppose $\lim_\limits{x\to0}\sin\frac{1}{x}=A$ where $A\in\mathbb R$.

The negation of the definition of limit is:

$\exists\epsilon>0$ such that $\forall\delta>0$, there is some $x\in\mathbb R$ such that $0<|x|<\delta$ and $\sin\frac{1}{x}\ge\epsilon$.

Let $\epsilon=\frac{1}{2}$. We want to define $x$ so that $|x|$ is less than $\delta$, but $\frac{1}{x}$ is $\frac{\pi}{2}+2\pi n$.

Let $x=\frac{1}{\frac{\pi}{2}+2\pi n}$ where $n$ is a sufficiently large integer so that $|x|<\delta$ (this statement may require more proof, but it is fairly obvious that you can pick such an $n$.

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

Voila. No need for cases.

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    $\begingroup$ You suppose by contradiction that $\lim_{x \rightarrow \infty} \sin \frac{1}{x} = A$, yet you write "The negation of the definition of a limit is...and $\sin \frac{1}{x} \ge \epsilon$." I have two questions: (1) where did the $A$ go; (2) why did you drop the absolute value bars? Shouldn't it read $|\sin \frac{1}{x} - A | \ge \epsilon$? $\endgroup$ – user193319 Feb 19 '17 at 0:20
  • $\begingroup$ Yeah this is not a good answer. This method shows that, if the limit exists, it must be $1$. A similar extension shows that it can't be $1$. $\endgroup$ – Elliot G Feb 19 '17 at 0:26
  • $\begingroup$ @ElliotG Regarding the statement you said would require further reasoning, could one say that the choice of $n \in \mathbb{N}$ is possible due to the archimedian property? $\endgroup$ – Lucas Jan 16 at 19:59
  • $\begingroup$ @ElliotG wouldn't it be better if the sequence $x_n$ is $\frac{1}{\frac{\pi}{2}+\pi n}$? Then $\sin{\frac{1}{x_n}}$ would be an alternating sequence of $1$ and $-1$ and thus, won't converge $\endgroup$ – Debartha Paul Feb 10 at 16:34

Warning: Case 1. Your claim that $$sin(1/x)\ is \ at \ most\ 1,\ then \ |\sin(1/x) - A| < |1-A|$$

requires a proof. It seems to me that it is not true.

  • $\begingroup$ If I had said "Since $\sin(1/x)$ is at most 1, then $|\sin(1/x) - A| \leq |1 - A|$", would it be correct? $\endgroup$ – Kurtland Chua Jul 10 '15 at 5:30
  • $\begingroup$ Nô it is not correct $\endgroup$ – Idris Addou Jul 10 '15 at 5:36
  • $\begingroup$ The reverse inequality is correct in this case 1 $\endgroup$ – Idris Addou Jul 10 '15 at 5:37
  • $\begingroup$ Since Case 1 relies on the use of $|1 - A|$, could I just state that in the interval $(-\delta,\delta), \exists x$ s.t. $|\sin(1/x) - A| = |1 - A|$? $\endgroup$ – Kurtland Chua Jul 10 '15 at 5:41

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