How to prove that the derivative of sin x is cos x using the limit definition?

Here is the complete geometric proof of both limits. **Theorem 1:** $\displaystyle \lim_{h \to 0} \frac{\sin h}{h} = 1$ **Proof (geometric):** Consider the unit circle. Let $h$ be a small positive angle in radians. Draw the following three areas: 1. Area of triangle $\triangle OAB$ where $A = (1,0)$ and $B = (\cos h, \sin h)$: $\frac12 \sin h$ 2. Area of sector $OAB$: $\frac12 h$ 3. Area of triangle $OAC$ where $C = (1, \tan h)$: $\frac12 \tan h$ From the diagram, for $0 < h < \pi/2$: $$\frac12 \sin h \leq \frac12 h \leq \frac12 \tan h$$ Multiplying by $2$: $$\sin h \leq h \leq \tan h = \frac{\sin h}{\cos h}$$ Dividing by $\sin h > 0$: $$1 \leq \frac{h}{\sin h} \leq \frac{1}{\cos h}$$ Taking reciprocals (reversing inequalities): $$\cos h \leq \frac{\sin h}{h} \leq 1$$ By the Squeeze Theorem, as $h \to 0^+$, $\cos h \to 1$, so $\frac{\sin h}{h} \to 1$. The same argument works for $h \to 0^-$ by symmetry. **Theorem 2:** $\displaystyle \lim_{h \to 0} \frac{\cos h - 1}{h} = 0$ **Proof:** \begin{align*} \lim_{h \to 0} \frac{\cos h - 1}{h} &= \lim_{h \to 0} \frac{(\cos h - 1)(\cos h + 1)}{h(\cos h + 1)} \\ &= \lim_{h \to 0} \frac{\cos^2 h - 1}{h(\cos h + 1)} \\ &= \lim_{h \to 0} \frac{-\sin^2 h}{h(\cos h + 1)} \\ &= -\lim_{h \to 0} \frac{\sin h}{h} \cdot \frac{\sin h}{\cos h + 1} \\ &= -1 \cdot \frac{0}{2} = 0 \end{align*} With both limits established, the derivative follows directly: \begin{align*} \frac{d}{dx}\sin x &= \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} \\ &= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\ &= \sin x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} + \cos x \cdot \lim_{h \to 0} \frac{\sin h}{h} \\ &= \sin x \cdot 0 + \cos x \cdot 1 = \cos x \end{align*}