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?
Best Answer
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 $\sin\frac{1}{x}=\sin(\frac{\pi}{2}+2\pi n)=1$.
Voilà. No need for cases.