<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] You can just sum-to-product it from the beginning and use $\lim_{x\rightarrow 0}\sin(x)/x = 1$:
\begin{align}
\lim_{x\rightarrow 0}\frac{\sin(x^2+2)-\sin(x+2)}{x} &= \lim_{x\rightarrow 0}\frac{2\cos[2+x(x+1)/2]\sin[x(x-1)/2]}{x}
\\ & = \cos(2)\lim_{x\rightarrow 0}\frac{\sin[x(x-1)/2]}{x/2}
\\ &= -\cos(2)\lim_{x\rightarrow 0}\frac{\sin[x(x-1)/2]}{x(x-1)/2} 
\\ &= -\cos(2)
\end{align}
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