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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)?

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    However, thank you for your opinion.2012-03-06

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