L'Hopital's rule and $\frac{\sin x}x$
It is a correct application of l'Hôpital's rule, but using it to prove that the derivative of $\sin x$ is $\cos x$ is possibly circular logic, depending on your definition of $\sin x$.
If we did not know what the derivative of $\sin x$ was, but we did know the angle sum formula, here are some steps we could take to compute that derivative, starting from first principles, i.e. the definition of the derivative.
$$ (\sin x)'= \lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h} = \lim_{h\to 0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h} \\ = \cos x\cdot\left(\lim_{h\to 0}\dfrac{\sin h}{h}\right)+ \sin x\cdot\left(\lim_{h\to 0}\dfrac{\cos h -1}{h}\right) $$
To complete the computation of the derivative of $\sin x$, we must first know the limit of $\sin h/h$ and $(\cos h-1)/h$. We can't use l'Hôpital at this point because to use l'Hôpital you need to know the derivative of $\sin x$ at $0$, which is the very thing we are trying to compute. Assuming what you're trying to prove is a logical error known as "begging the question". Mathematics is axiomatic, so that every result builds on previous results, and no arguments are circular.
But if we can confirm the derivative of $\sin x$ by some other means, for example by defining it in terms of its Taylor series or a differential equation, then we may use it in a l'Hôpital computation to derive $\lim \sin x/x$ without fear of being circular. While not logically incorrect, this would still be fairly redundant: if you know the derivative of $\sin x$ just observe that, by definition, $\lim_{h\to 0}\sin x/x$ is just the derivative at 0. Using the machinery of l'Hôpital is overkill to get an answer you already know.
However if you're in a situation where you don't care about logical foundations and rigor, you're not proving theorems from first principles about calculus of trigonometric functions, you just need to compute $\lim\sin x/x$ and want to allow all known formulas and techniques, feel free to use l'Hôpital. It is correct.
I expand my comment to ziggurism's answer here.
Suppose that by a lot of hardwork, patience and ... (add more nice words if you like) I have obtained these three facts:
- $\cos x$ is continuous.
- $\dfrac{d}{dx}(\sin x) = \cos x$
- L'Hospital's Rule
and my next goal is establish the following fact: $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{*}$$ Because of all that hardwork, patience and ... I know that my goal $(*)$ is an immediate consequence of just the fact $(2)$ alone stated above. Then why would I combine all the three facts mentioned above to achieve my goal? To borrow an idea from user "zhw.", wouldn't a person doing this would be considered silly?
More often than not, many students don't really understand what's going behind the scenes when we use the mantra of "differentiate and plug" championed by L'Hospital's Rule. The act of differentiation itself entails that we know certain limits (and rules of differentiation) and further most of the derivatives are continuous (so that plugging works after differentiation step).
If one is so fond of L'Hospital's Rule why not put that to a better use to solve complex problems (like this and this) instead of using it to obtain limits which are immediate consequences of differentiation formulas.
How do you prove that $\frac{d}{dx}[\sin(x)]=\cos(x)$? If you use the limit $$\lim_{x\to 0}\frac{\sin(x)}{x}=1 \qquad\qquad(1.1)$$ than you can't use the L'Hopital rule, because it is an example of circular logic.
You can avoid it, definying $$\sin(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\ \ \ \ \forall x\in\mathbb{R}$$ and $$\cos(x)=\sum_{n=0}^{+\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}\ \ \ \forall x\in\mathbb{R}$$.
Once you proved that $\frac{d}{dx}[\sin(x)]=\cos(x)$ and $\cos(0)=1$ than you can use the l'Hopital rule to calculate the limit (1.1).