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Prove using the definition of a limit that $$\lim_\limits{x\to0}\sin(x)=0. $$ And $$\lim_\limits{x\to0}\frac{\sin(x)}{x}=1.$$

I need to prove that $\forall\epsilon>0 \,\,\exists \delta >0$ that $$\left|\frac{\sin(x)}{x}-1\right|<\epsilon\quad \text{whenever} \quad |x|<\delta.$$

So $$\left|\frac{\sin(x)}{x}-1\right|=\left|\frac{\sin(x)-x}{x}\right|< \frac{1+|x|}{|x|}=\frac1{|x|}+1<\epsilon $$ So in this case would my $\delta $ be $1/(\epsilon-1)$? But then this would be true for $|x|>\delta$, and it might be negative, so how should I do this?

And for the first: $$|\sin(x)|<1<\epsilon$$ But how would this help me? Since I need it in the form $|x|<f(\epsilon)$ where $f$ is any expression of $\epsilon$.

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I know that $\sin(x)$ behaves like $x$ for values close to $0$ but using this argument for the proofs above seems kind of circular.

Also, I am familiar with the geometric proof for the second limit, but I wanted to know how an epsilon-delta proof might look like if I did not have any knowledge of such proof. I should mention that methods more advanced than taking a limit (ex: Taylor series) should not be used.

user372003
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  • This might help. – drhab Mar 01 '18 at 09:06
  • use the Taylor theorem – Masacroso Mar 01 '18 at 09:08
  • @drhab I am familiar with the geometric proof, I was looking for something that did not need to have such knowledge – user372003 Mar 01 '18 at 09:08
  • @Masacroso I am guessing that in this case it is assumed to not use methods more advanced than limits. – user372003 Mar 01 '18 at 09:10
  • @user372003 I can see only two approach: a geometrical proof or an analytical proof (using Taylor of analytic definition of sine analytic definition) – Masacroso Mar 01 '18 at 09:15
  • You want to avoid using that $1$ is an upper bound for $\sin$ here, since an important part of this convergence is that $\sin$ and $x$ are both going to $0$. Squeeze theorem might be a good method here. – Alex Jones Mar 01 '18 at 09:16
  • @AlexanderJ93 Squeeze theorem? How would that be a good method? I mean it is easy to apply it to the case where $x$ is approaching infinity, but I can’t see how it would make things simpler for this case. – user372003 Mar 01 '18 at 09:25
  • If you can show that, within a certain range of $0$, that $\cos(x) \leq \sin(x)/x \leq 1$, then you should have it. Showing that $\cos(x) \rightarrow 1$ as $x \rightarrow 0$ is a much simpler limit. – Alex Jones Mar 01 '18 at 09:27
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    The essential question here is the following: What is your definition of $\sin$? – Christian Blatter Mar 01 '18 at 09:28

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