Given a function $f\in(-a,a)\smallsetminus\{0\} \to \mathbb{R}$. Prove that if $\lim_{x\to 0} f(x)=L$ then $\lim_{x\to 0}f({\sin(x))=L}$

Question

Given a function $f\in(-a,a)\smallsetminus\{0\} \to \mathbb{R}$. I want to prove that if $\lim_{x\to 0} f(x)=L$ then $\lim_{x\to 0}f({\sin(x))=L}$.

I was trying to prove it using Cauchy's definition but got stuck along the way.

Any help would be appreciated.

There is no logical connection in your question's first part. Perhaps you mistyped something. Check again carefully your question... — DonAntonio, Dec 25 '17 at 13:00
It is still not true; a counterexample would be $a=1$, $f(x)=x+1$, and $L=1$. — hmakholm left over Monica, Dec 25 '17 at 13:25
https://math.stackexchange.com/questions/784985/proving-theres-a-function-f-such-that-lim-x-to-0fx-l-rightarrow-lim?rq=1 — Nosrati, Dec 25 '17 at 13:29
@MyGlasses I have seen this thread but they don't prove the general case there — Lin Tanaka, Dec 25 '17 at 13:30
This is nothing but substitution in limits. You can use the more general result described in this answer. — Paramanand Singh, Dec 25 '17 at 15:11

score 2 · Answer 1 · answered Dec 25 '17 at 13:49

2

Take any $\;\epsilon>0\;$, then

$$\begin{cases} I\;\;\lim\limits_{x\to0}f(x)=L\implies \exists\,\delta_1>0\,\,s.t.\,\, |x|<\delta_1\implies|f(x)-L|<\epsilon\\{}\\ II\;\;\lim\limits_{x\to0}\sin x=0\implies \exists\delta_2>0\,\,s.t.\,\,|x|<\delta_2\implies |\sin x|<\delta_1\end{cases}$$

Define now $\;\delta:=\min(\delta_1,\,\delta_2)\;$ , then

$$|x|<\delta\stackrel{II}\implies |\sin x|<\delta_1\stackrel I\implies |f(\sin x)-L|<\epsilon$$

answered Dec 25 '17 at 13:49

DonAntonio

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BTW, no need of that $\delta;$ there. It is enough to take $;|x|<\delta_2;$ and we're done. – DonAntonio Dec 25 '17 at 14:02

score 1 · Answer 2 · answered Dec 25 '17 at 13:48

Let $\varepsilon>0$. Take $\delta>0$ such that$$|x|<\delta\implies\bigl|f(x)-L\bigr|<\varepsilon.$$Then $|x|<\delta\implies\bigl|\sin(x)\bigr|<\delta$, because$$(\forall x\in\mathbb{R}):\bigl|\sin(x)\bigr|\leqslant|x|,$$and therefore$$|x|<\delta\implies\bigl|f\bigl(\sin(x)\bigr)-L\bigr|<\varepsilon.$$

Shashi · Answer 3 · 2017-12-25T16:30:36.973

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For any sequence $x_n\neq 0$ such that $x_n\to 0$ we have $f(x_n) \to L$. That is just the assumption reformulated.

Take $x_n=\sin(y_n) $ such that $y_n \in(-\pi, \pi)\setminus \{0\} $ and $y_n\to 0$. This $y_n$ is arbitrary enough for this setting (why? $\star$). We have $x_n\to 0$ by the continuity of $\sin(\cdot) $. And so $f(\sin(y_n)) =f(x_n)\to L$. Since $y_n$ was arbitrary we have: $$\lim_{x\to 0}f(\sin(x))=L$$

$(\star) $ The answer for "why?"

Take an arbitrary $z_n\neq 0$ such that $z_n\to 0$. There exist $N$ such that for all $n\geq N$ we have $z_n\in (-\pi, \pi) \setminus \{0\} $. For the limit it is enough to consider $y_{n} =z_{N+n} $. So $y_n$ is arbitrary enough.

edited Dec 25 '17 at 16:30

answered Dec 25 '17 at 13:57

Shashi

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Well, $;y_n;$ isn't that arbitrary as you chose them from $;(-\pi,\pi)\setminus{0};$ (who knows why, btw...) – DonAntonio Dec 25 '17 at 14:04
@DonAntonio take $z_n\neq 0$ such that $z_n\to 0$. There exist $N$ such that for all $n\geq N$ we have $z_n\in (-\pi, \pi) \setminus {0} $. For the limit it is enough to consider $y_{n} =z_{N+n} $. And that is why I said so. Now you agree? – Shashi Dec 25 '17 at 14:25
Oh, I think I know what you tried to do...but perhaps it is worthwhile to edit your answer and add a little word about this. – DonAntonio Dec 25 '17 at 16:27
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@DonAntonio done that! Thanks for your comment! – Shashi Dec 25 '17 at 16:30

score 0 · Answer 4 · answered Dec 25 '17 at 14:08

0

Let t=sin(x). If x approaches 0 then t approaches 0, therefore f(t)=f(sin(x)) approaches L.

answered Dec 25 '17 at 14:08

Mohammad Riazi-Kermani

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Given a function $f\in(-a,a)\smallsetminus\{0\} \to \mathbb{R}$. Prove that if $\lim_{x\to 0} f(x)=L$ then $\lim_{x\to 0}f({\sin(x))=L}$

4 Answers4