You did indeed miss a simplification:
$$\lim \limits_{x \to 0} \space {(-x\sin{x} + \cos{x}) - \cos{x} \over (x\cos{x} + \sin{x})}=\lim \limits_{x \to 0} \space {-x\sin{x}\over (x\cos{x} + \sin{x})}$$
It's not a lot simpler, but it is simpler.
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.
Best Answer
$$\lim_{x\to 0}\frac{\tan(\beta x)-\beta\tan(x)}{\sin(\beta x)-\beta\sin(x)}$$ yields an indeterminate $\lim_{x\to 0}\frac{0}{0}$ case. Hence $$\lim_{x\to 0}\frac{\tan(\beta x)-\beta\tan(x)}{\sin(\beta x)-\beta\sin(x)}=\lim_{x\to 0}\frac{\frac{d}{dx}(\tan(\beta x)-\beta\tan(x))}{\frac{d}{dx}(\sin(\beta x)-\beta\sin(x))}=\lim_{x\to 0}\frac{\beta \sec^2(\beta x)-\beta \sec^2(x)}{\beta\cos(\beta x)-\beta\cos(x)}=\lim_{x\to0}\frac{\frac{1}{\cos^2(\beta x)}-\frac{1}{\cos^2(x)}}{\cos(\beta x)-\cos(x)}=\lim_{x\to 0}\frac{\cos^2(x)-\cos^2(\beta x)}{\cos^2(x)\cos^2(\beta x)}\cdot\frac{1}{\cos(\beta x)-\cos(x)}$$ $$=\lim_{x\to0}-\frac{\cos(\beta x)+\cos(x)}{\cos^2(x)\cos^2(\beta x)}=-\frac{\cos(0)+\cos(0)}{\cos^2(0)\cos^2(0)}=-2$$