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The question is :

Show that, $$\lim_{x \to 0} \frac {\sin \frac {1} {x}} {\sin \frac {1} {x}}$$ does not exist.

How to solve it by the defination of limit? Can it be solved using sequential criterion? Please help me. Thank you in advance.

Michael Hardy
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    Does the limit really not exist? Doesn't the expression equal $1$ for all non-zero $x$? – Brian Tung Jul 25 '16 at 17:36
  • That limit is 1. – Meadara Jul 25 '16 at 17:39
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    We should qualify that we must have $x$ such that $\sin(1/x) \neq 0$. Otherwise, the ratio is undefined. But, for every sequence $x_{1}, x_{2}, \ldots$, that increases without bound and satisfies $\sin(1/x_{k}) \neq 0$, the limit of your ratio will be $1$ as $k \rightarrow +\infty$. – avs Jul 25 '16 at 17:40
  • But in my book it is given that the limit does not exist. –  Jul 25 '16 at 17:42
  • Are you sure @avs , about the case x=0 , because it is the same curve so it should have the same value for all x even at x=0. – Meadara Jul 25 '16 at 17:42
  • What will happen if $sin \frac {1} {x} = 0$? –  Jul 25 '16 at 17:45
  • it will still be 1 because it is the same curve. – Meadara Jul 25 '16 at 17:45
  • The case $x=0$ does not bother me as much as the case $x = 1 / \pi$, which turns the ratio into the indeterminacy $0/0$. – avs Jul 25 '16 at 18:15

3 Answers3

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You wanted an $\varepsilon-\delta$ proof, but the claim is wrong. So I will provide an $\varepsilon-\delta$ that the claim is wrong. More specifically, I will show that $$\lim_{x \to 0} \frac {sin \frac {1} {x}} {sin \frac {1} {x}}=1$$ Given $\varepsilon>0$, put $\delta=\varepsilon$. Then whenever $0<|x-0|<\delta$ (and $1\neq x\pi k$ for some $k\in\mathbb{Z}$), we have $$\left| \frac {sin \frac {1} {x}} {sin \frac {1} {x}} -1\right|=0<\varepsilon.$$

Aweygan
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  • In fact $x = \frac {1} {{\pie}{k}}$ does not belong to the domain of defination of the given function, for any $k \in \mathbb {Z}$. Isn't it? –  Jul 25 '16 at 18:06
  • That's true. You could replace "for some $k\in\mathbb{Z}$" in my answer with "for all $k\in\mathbb{Z}$" – Aweygan Jul 25 '16 at 18:09
  • Sorry for my wrong claim. My book misguides me badly. –  Jul 25 '16 at 18:13
  • No apologies are necessary. I would recommend that you double check the claims in your book, now that you've found some errors. This wouldn't be the first time a book had an error in it. – Aweygan Jul 25 '16 at 18:17
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Yeah for all $x\in\mathbb R$, and $x\ne0$, $$\lim_{x \to 0} \frac {\sin \frac {1} {x}} {\sin \frac {1} {x}} =1$$ since it became a constant function!

Michael Hardy
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Sathasivam K
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Let $f(x)$ be an abbreviation for $(\sin 1/x)/\sin 1/x .$ The statement $$\forall e>0\;\exists d>0\;\forall x \;(|x|<d\implies |f(x)-1|<e)$$ is FALSE. Because for any $d>0$ there exists $x\in (-d,d)$ at which $f(x)$ does not exist :

Given $d>0$ : For any large enough $n\in \mathbb Z^+$ we have $0<1/\pi n <d $, so if $x=1/\pi n$ then $x\in (-d,d) ,$ and $\sin 1/x=0$ so the formal expression $(\sin 1/x)/\sin 1/x$ is $0/0.$

DanielWainfleet
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  • Your statement is false. It should be $\forall e>0;\exists d>0;\forall x\in\textrm{Dom}(f) ;(|x|<d\implies |f(x)-1|<e)$. We need to take the domain into account – Aweygan Jul 25 '16 at 18:26
  • The Q does not specify that x is restricted to any particular proper subset of R. That is precisely the point of the Q. – DanielWainfleet Jul 25 '16 at 19:25