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I was wondering is there another way of proving that limit of $sin(x)/x = 1$ as $x$ approaches $0$ without using l'hospital rule but using equivalence relations. I was reading a calculus book and i see that there some functions that are equivalence to one another like $sin(x)$ is equivalent to $x$.

user146269
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    Well, (depending on what equivalence you're talking about) that would be circular argument. We say that $\sin x \sim x$ exactly because $\lim\limits_{x\to 0}\frac{\sin x}{x} = 1$. – Eff Mar 12 '15 at 15:53
  • What does your book mean that $\sin x$ is equivalent to $x$? – MT_ Mar 12 '15 at 15:55
  • By 'equivalence relation' do you mean the fact that ($\alpha(x),\alpha_1(x),\beta(x),\beta_1(x)$ having limits $0$ when $x\to a$): $\lim_{x\to a}\frac{\alpha(x)}{\alpha_1(x)}=1, \lim_{x\to a}\frac{\beta(x)}{\beta_1(x)}=1\implies \lim_{x\to a}\frac{\alpha(x)}{\beta(x)}=\lim_{x\to a}\frac{\alpha_1(x)}{\beta_1(x)}$? – user26486 Mar 12 '15 at 16:04
  • More like using little $o$ and big $O$ as equivalence relations – user146269 Mar 15 '15 at 20:20

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I wouldn't go so far as to say that any "equivalence" that $\sin(x)$ and $x$ have is useful here. The ratio of their limit at $0$ is $1$, but other than that, they're very different functions. If your definition of "equivalent" is that the ratio of their limit at $0$ is $1$, then you can't use equivalence in your proof, because that would be circular.

You can prove this without any advanced techniques using the Squeeze Theorem. In particular, when $x$ is close to $0$, $\frac{\sin(x)}{x}$ can be squeezed between $cosx$ and $1$. See this PDF for a fuller explanation.

NoName
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