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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. $\endgroup$ Jul 10 '15 at 2:27
  • $\begingroup$ Do you know Heine's Theorem? $\endgroup$
    – Vim
    Jul 10 '15 at 2:33
  • $\begingroup$ Just as a warning to any readers, I believe the inequality in your case 1 is invalid: $|\sin(1/x) - A| < |1-A|$. Consider when $\sin(1/x)$ is a negative value and $A$ is any positive value (for case 1, this is restricted to "greater than $1$"). $\endgroup$
    – S.Cramer
    May 30 at 19:48
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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.

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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
    Feb 10 at 16:34
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Warning: Case 1. Your claim that $$\sin(1/x)\ \text{is} \ \text{at} \ \text{most}\ 1,\ \text{then} \ \left|\sin\left(\frac1x\right) - A\right| < |1-A|$$

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

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  • $\begingroup$ If I had said "Since $\sin(1/x)$ is at most 1, then $|\sin(1/x) - A| \leq |1 - A|$", would it be correct? $\endgroup$ Jul 10 '15 at 5:30
  • $\begingroup$ Nô it is not correct $\endgroup$ Jul 10 '15 at 5:36
  • $\begingroup$ The reverse inequality is correct in this case 1 $\endgroup$ 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. $|\sin(1/x) - A| = |1 - A|$? $\endgroup$ Jul 10 '15 at 5:41
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Although there are several less involved ways of doing this, if you want to break this proof into your three cases, you can proceed by using the negated version of the definition.

To do this, you will need a few lemmas:

i) The Archimedean Property - for our purpose, we can frame it as $\exists n \in \mathbb N ( 2\pi n \gt \frac{1}{\delta})$.

ii) For $a,b \gt 0$, we have that $a \gt b \rightarrow \frac{1}{a} \lt \frac{1}{b}$. Applying this to our above statement, we have: $\frac{1}{2 \pi n} \lt \delta$

iii) For any $z \in \mathbb R$, $\sin(z)$ is defined. In particular, we have $\forall n \in \mathbb N [ \sin(2 \pi n) = 0]$

iv) Several variants of the Triangle Inequality

v) For any $z \in \mathbb R$, $|\sin(z)| \leq 1$.


For each case, we want to construct an $\epsilon^* \gt 0$ such that the following statement is true:

$\forall \delta \gt 0 \ \exists x^* \in \mathbb R \Big [0 \lt |x^*| \lt \delta \rightarrow |\sin(\frac{1}{x^*})-A| \geq \epsilon^*\Big]$.

In English, for any arbitrary $\delta$, we want to show that there is at least one $x^*$ within the interval of $(-\delta, \delta)$ such that $|\sin(\frac{1}{x^*})-A| \geq \epsilon^*$.

Case 1: $|A| \gt 1$

Let $\epsilon^* = \frac{|A|-1}{2}$, which we note is greater than $0$.

$$\Big\lvert\sin(\frac{1}{x})\Big\rvert\leq1 \implies -\Big\lvert\sin(\frac{1}{x})\Big\rvert \geq -1 \implies |A| -\Big\lvert\sin(\frac{1}{x})\Big\rvert \geq |A| -1 \gt \frac{|A|-1}{2} \gt 0$$

Therefore:

$$\Big\lvert \sin(\frac{1}{x})-A \Big\rvert = \Big\lvert A - \sin(\frac{1}{x}) \Big\rvert \geq |A| -\Big\lvert\sin(\frac{1}{x})\Big\rvert \gt \frac{|A|-1}{2} = \epsilon^*$$

So for this case, we could literally choose any value so long as it satisfies $0 \lt |x^*| \lt \delta$. For simplicity, choose $x^*=\frac{\delta}{2}$

Case 2: $|A|=1$

Let $\epsilon^*=\frac{1}{2}$.

So we need to prove: $\forall \delta \gt 0 \ \exists x^* \in \mathbb R \Big [0 \lt |x^*| \lt \delta \rightarrow |\sin(\frac{1}{x^*})-A| \geq \frac {1}{2}\Big]$.

For an arbitrary $\delta$, choose an $n \in \mathbb N$ such that $x^*=\frac{1}{2\pi n} \lt \delta$. Note that $\sin\Big(\frac{1}{\frac{1}{2\pi n}}\Big)=\sin (2 \pi n)=0$

For such an $x^*$, our expression $|\sin(\frac{1}{x^*})-A|$ simplifies to $|0 - A| = 1$, and $1 \geq \frac{1}{2}=\epsilon^*$.

Case 3: $|A| \lt 1$

You can repeat the argument exactly as Case 2, but let $\epsilon^*=\frac{|A|}{2}$.

You will arrive at $|0 - A|= |A| \geq \frac{|A|}{2} = \epsilon^*$

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