<s>[INST] Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $ 
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$ 

I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
 $$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
 $$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
 [/INST] First notice that
$$\lim_{x\to 0} \frac{\cos x - 1}{x} = \lim_{x\to 0} \frac{2\sin^2\frac{x}2}{x} = \lim_{x\to 0} \left(\frac{\sin^2\frac{x}2}{\frac{x}2}\right)\cdot\sin\frac{x}2 = 1 \cdot 0 = 0$$
so
$$\lim_{x\to 0}\frac{ \cos{x^2}-\cos{x}}{x} = \lim_{x\to 0} \left(\frac{\cos x^2-1}{x^2}\right)x + \lim_{x\to 0} \frac{1-\cos x}{x} =  0$$
Finally we get
$$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} =\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x}  = -\cos 2$$
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