With the linear approx. of $f(x)= sin(x)$ around $0$ Calculate $\lim_{\theta\to 0} \frac{sin\theta}{\theta}$ With the linear approximation of $f(x)= sin(x)$ around $0$, calculate:
$$ \lim_{\theta\to 0} \frac{\sin\theta}{\theta}$$
Figured I have to use L'Hospital's Rule, but I think I don't get how to calculate the derivative of theta.
$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta}=\lim_{\theta \to 0} \frac{\frac{d}{d\theta}\sin\theta}{\frac{d}{d\theta}\theta}=\lim_{\theta \to 0} \frac{\cos \theta}{1}=\frac{\cos 0}{1}=1$$

Two ways to go about doing this:
Taylor Series
We can use the Taylor expansion of $\sin(x)$
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$
$$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^3}{5!}-\frac{x^6}{7!}+\ldots$$
So $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$
Squeeze Theorem

The area of the triangle in the image above is $\displaystyle \frac{\sin\theta}{2\cos\theta}$.
The area of the small circular sector is $\displaystyle \frac{\theta}{2}$.
The area of the big circular sector is $\displaystyle \frac{\theta}{2\cos^2\theta}$.
So, for $0<\theta<\displaystyle \frac{\pi}{2}$, we have
$$\displaystyle \frac{\theta}{2}<\displaystyle \frac{\sin\theta}{2\cos\theta}<\displaystyle \frac{\theta}{2\cos^2\theta}$$
Since $\displaystyle \frac{2\cos\theta}{\theta}$ is positive for $0<\theta<\displaystyle \frac{\pi}{2}$, we can multiply by $\displaystyle \frac{2\cos\theta}{\theta}$:
$$\cos\theta<\frac{\sin\theta}{\theta}<\frac{1}{\cos\theta}$$
Since $\displaystyle \lim_{\theta \to 0} \cos\theta=\lim_{\theta \to 0} \frac{1}{\cos\theta}=1$, we can use Squeeze Theorem to conclude $\displaystyle \lim_{\theta \to 0}\frac{\sin\theta}{\theta}=1$.
