삼각함수의 미분

First, we note that our function is continuous, because $$\lim_{x \to 0} f(x)=\lim_{x \to 0} \frac{\sin x}{x}=1=f(0)$$ This is a handy thing to check: if the function were discontinuous at $x=0$, then we would automatically know that it was not differentiable there.

Now, on to the derivative. We can use the limit definition:

$$\begin{aligned} f'(0)&=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}\qquad\text{ if it exists}\\ &=\lim_{h \to 0} \frac{f(h)-1}{h}\\ &=\lim_{h \to 0} \frac{\frac{\sin h}{h}-1}{h} \end{aligned}$$

As $h$ approaches 0, both the numerator and the denominator approach 0. So, to evaluate the limit, we need to do more work. The key insight we can use is a result that was shown in the text while evaluating the derivative of sine. When $h$ is close to 0, $\cos h \leq \frac{\sin h}{h} \leq 1$. We use this to bound our limit, and then apply the squeeze theorem.

$$\begin{aligned} &\cos h&&\leq&&\frac{\sin h}{h}&&\leq&&1\\ \text{So, }\qquad&\cos h-1&&\leq&&\frac{\sin h}{h}-1&&\leq&& 1-1=0 \end{aligned}$$

and $\frac{1}{h}\left(\frac{\sin h}{h}-1\right)$ is between $0$ and $\frac{\cos h-1}{h}$. We can evaluate the limit of $\frac{\cos h-1}{h}$, by noticing its similarity to the definition of the derivative.

$$\begin{aligned} \lim_{h \to 0}\frac{\cos h-1}{h}&=\lim_{h \to 0}\frac{\cos(0+h)-\cos(0)}{h}\\ &=\frac{\mathrm{d}}{\mathrm{d}x}\left.\left\{\cos x \right\}\right|_{x=0}=0 \end{aligned}$$

So, by the Squeeze Theorem, $f'(0) =\displaystyle \lim_{h \to 0}\frac{\frac{\sin h}{h}-1}{h}=0$.