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So I’m not sure how to do this problem, I almost understand the squeeze theorem, but I have no idea how to prove it using the provided equation!

I would be extremely thankfull if you could help me:)

Using $\;\left|1-\dfrac{\sin(x)}x\right|<\dfrac{x^2}2\,$ prove that $\,\lim\limits_{x\to0}\dfrac{\sin(x)}x=1\,.$

Antanas Strumskis
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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Feb 06 '23 at 19:27
  • The squeeze theorem isn't entirely necessary here. You could just do a direct $\delta-\epsilon$ proof. – aschepler Feb 06 '23 at 19:52
  • Hint: fixing $\epsilon>0$, what happens if $|x|<\delta$ with $\delta:=\sqrt{2\epsilon}$? – J.G. Feb 06 '23 at 19:53
  • @J.G. why giving advice to root square epsilons, $|x|<1\implies x^2<|x|$ and we can deal with normal epsilons. When $x\to 0$ any polynomial in $x$ can be reduced to $|p(x)|< c|x|$ (c=sum of abs coeffs) with the same idea that $|x^n|<|x|$ for $|x|<1$. – zwim Feb 06 '23 at 23:03
  • @zwim Fair point. – J.G. Feb 06 '23 at 23:08

1 Answers1

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If you want to apply the squeeze theorem:

$$ 0 \leq |1-\sin(x)/x| \leq x^2/2$$

If we let $x \rightarrow 0 $ then the RHS tends to $0$ and so the squeeze theorem states that $|1-\sin(x)/x| \rightarrow 0$ as $x \rightarrow 0$, meaning that $\sin(x)/x \rightarrow 1$ as $x \rightarrow 0$.

Tanamas
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  • Thank you for your answer, but i dont know what "RHS" means. Is there any way you could dumb the whole equation down for me? – Antanas Strumskis Feb 07 '23 at 12:05
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    Absolutely, RHS means "right hand side". Look at $|1-\sin(x)/x|$ in the first expression which you've also been given in the problem. Since it's in absolute values, it's certainly positive, so it's bounded below by $0$. We also know from the problem that the middle expression is less than $x^2/2$. So if we let $x\rightarrow 0$, the $|1-\sin(x)/x|$ is squeezed between $0$, so it has to be $0$ according to the squeeze theorem. But that can only be true if $1=\sin(x)/x$ for when $x \rightarrow 0$. – Tanamas Feb 07 '23 at 16:46
  • Thanks you for taking the time! You're very helpfull:) – Antanas Strumskis Feb 07 '23 at 17:21
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    @AntanasStrumskis No worries! Good luck on your studies :) – Tanamas Feb 07 '23 at 17:35