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I want to prove the following limit \begin{equation} \lim_{x\rightarrow 0}\frac{\sin x}{x}=1 \end{equation} using the definition of limit. Then: \begin{equation} \left| \frac{\sin x}{x}-1 \right|=\left|\frac{\sin x}{ x}\right| \left|1 -\frac{x}{\sin x} \right|\leq \left|1 -\frac{x}{\sin x} \right| \end{equation} Also \begin{equation} \left|\frac{x}{\sin x}-1 \right|\leq \left|\frac{1}{\cos x}-1 \right|=\left|\frac{\sin^{2} x}{\cos x(1+\cos x)}\right| \end{equation} noting that $\cos x>0$ in the neighbourhood $(-\delta_{\epsilon},\delta_{\epsilon})$, I have: \begin{equation} \leq \left|\frac{\sin^{2} x}{\cos x}\right|=\left| \sin x \right| \left| \tan x \right|\leq \left| \tan x \right|\leq \tan \delta_{\epsilon} \leq \epsilon \end{equation} it is verified for $\delta_{\epsilon}=arc\tan(\frac{\varepsilon}{2})$ (for example) Can anyone "see" a simpler solution (using the definition of limit)?

Mark
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    How about this: If you accept that cosx is the derivative of sinx, then:

    The limit as h->0

    $\frac{sinh-sin0}{h}=cosh=1$

     – AQP Mar 06 '12 at 06:27
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    that is probably circular and certainly not in spirit of problem (to invoke derivatives) – ShawnD Mar 06 '12 at 06:57
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    Would you accept the "squeeze" principle? (i.e., if $f(x) \leq g(x) \leq h(x)$ for all $x \neq p$ in some neighborhood $N(p)$ and the limits of $f(x)$ and $h(x)$ as $x \to p$ both $= a$ then the limit of $g(x)$ as $x \to p$ is also $= a$?) If so, you can use $f(x) = \cos x$ and $h(x) = 1/\cos x$. –  Mar 06 '12 at 07:25
  • is a good solution, but I wanted to prove the $\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$ using the definition of limit. – Mark Mar 06 '12 at 07:32
  • Understood. I was just thinking the squeeze principle is pretty close to following directly from the definition, so figured it was worth suggesting that approach. I don't have a ready improvement to suggest to yours for directly from the definition. –  Mar 06 '12 at 07:36
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    See this http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X for 3 different proofs of the theorem. – Listing Mar 06 '12 at 10:26
  • However, thank you for your opinion. – Mark Mar 06 '12 at 18:44

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