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.
$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta}=\lim_{\theta \to 0} \frac{\frac{d}{d\theta}\sin\theta}{\frac{d}{d\theta}\theta}=\lim_{\theta \to 0} \frac{\cos \theta}{1}=\frac{\cos 0}{1}=1$$
 A: Others have already pointed out that you simply use that for $\theta$ small $\sin(\theta) \approx \theta$ and you can use this to calculate the limit. Only for completion of this answer, you then have
$$
\lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} = \lim_{\theta\to 0} \frac{\theta}{\theta} = \lim_{\theta \to 0} 1 = 1.
$$
I wanted to give also a quick comment on your use of L'Hopital's Rule:

(I have included your original image since)
It is not (in general) correct that
$$
\lim_{x\to 0} \frac{f(x)}{g(x)} = \color{red}{\lim_{x\to 0}}\frac{d}{dx} \frac{f(x)}{g(x)}.
$$
(You, of course, forgot the limit in front of the fractions.)
What is true, in your situation, is that
$$
\lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x\to 0} \frac{f'(x)}{g'(x)}.
$$
You have to remember to take the derivative of the top and the bottom separately. It looks like this is actually what you end up doing, but the notation is important. Using incorrect notation can cause a lot of confusion (In particular when grading written work and the teacher is tired :))
A: 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 $\displaystyle \lim_{\theta \to 0}\frac{\sin\theta}{\theta}=1$.
A: It says to use the linear approximation of $\sin \theta$ around $\theta=0$ which is $\sin \theta \approx \theta$. Therefore, we have:
$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta}=\lim_{\theta \to 0} \frac\theta\theta=1$$
A: The hint tells you to use the linear approximation of $\sin(x)$ near $0$.
For $x$ near $0$, $\sin(x) \approx x.$ So for $x$ near $0$,
$$\frac{\sin(x)}{x} \approx \frac{x}{x} = 1.$$
Thus $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$.
A: Well, the approximation of functions around $0$ can be obtained from Maclaurin series. 
Here, you have $sin(x)=x-1/6x^3+...$ as its Maclaurin series, 
and so its linear approximation is just the first term, i.e, $sin(x)\approx x$.
As a result, the answer would be $1$.
