sin(x)导数推导疑问：寻求非洛必达法则的极限求解方法

Great question—using L’Hospital’s Rule here is totally circular, since we’re trying to prove the derivative of $\sin(x)$ itself! Let’s walk through rigorous, non-circular ways to solve both limits using geometric intuition and basic calculus theorems.

1. $\lim_{h\to 0} \frac{\sin(h)}{h} = 1$ #

We can use a geometric argument with the unit circle and the squeeze theorem (also called the sandwich theorem) here:

• Imagine the unit circle centered at the origin. For a small positive angle $h$ (in radians, $0 < h < \pi/2$), let $A = (1,0)$, $B = (\cos h, \sin h)$, and consider the sector $OAB$, triangle $OAB$, and triangle $OAC$ (where $C = (1, \tan h)$).
• The area of triangle $OAB$ is $\frac{1}{2} \sin h$.
• The area of sector $OAB$ is $\frac{1}{2} h$ (since sector area is $\frac{1}{2} r^2 \theta$, and $r=1$ for the unit circle).
• The area of triangle $OAC$ is $\frac{1}{2} \tan h$.

These areas satisfy the inequality:
$$\frac{1}{2} \sin h < \frac{1}{2} h < \frac{1}{2} \tan h$$

Multiply through by 2 and divide by $\sin h$ (which is positive for $0 < h < \pi/2$, so the inequality directions stay the same):
$$1 < \frac{h}{\sin h} < \frac{1}{\cos h}$$

Take the reciprocal of all terms (this reverses the inequalities):
$$\cos h < \frac{\sin h}{h} < 1$$

As $h \to 0$, $\cos h \to 1$. By the squeeze theorem, $\frac{\sin h}{h}$ must approach 1 too. For negative $h$, note that $\sin(-h) = -\sin h$, so $\frac{\sin(-h)}{-h} = \frac{\sin h}{h}$, meaning the left-hand limit equals the right-hand limit. Thus, $\lim_{h\to 0} \frac{\sin(h)}{h} = 1$.

2. $\lim_{h\to 0} \frac{\cos(h) - 1}{h} = 0$ #

We can solve this with algebraic manipulation and the limit we just proved:

• Start by rationalizing the numerator:
  $$\frac{\cos h - 1}{h} = \frac{(\cos h - 1)(\cos h + 1)}{h(\cos h + 1)}$$
• Use the Pythagorean identity $\cos^2 h - 1 = -\sin^2 h$ to simplify the numerator:
  $$\frac{-\sin^2 h}{h(\cos h + 1)} = -\frac{\sin h}{h} \cdot \frac{\sin h}{\cos h + 1}$$
• Now take the limit as $h \to 0$:
  • We already know $\lim_{h\to 0} \frac{\sin h}{h} = 1$
  • $\lim_{h\to 0} \frac{\sin h}{\cos h + 1} = \frac{0}{1 + 1} = 0$

Multiplying these results gives:
$$\lim_{h\to 0} \frac{\cos h - 1}{h} = -1 \cdot 0 = 0$$

Putting It All Together #

Substitute these limits back into your derivative expression:
$$\frac{d}{dx} \sin(x) = \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x)$$

No circular reasoning required—just good old geometry and algebra!

内容的提问来源于stack exchange，提问作者user525966