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
 $$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$

The second limit can be evaluated as follows:
$\quad \quad \quad \lim_{x\to0} \frac{sin2(cosx^2 - cosx)}{x}$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2 - cosx}{x})$
$\quad \quad \quad \lim_{x\to0}  sin2 \cdot (\frac{cosx^2}{x} - \frac{cosx}{x})$
$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{1}{x} \cdot (cosx^2 - cosx))$
$\quad \quad \quad sin2 \cdot \lim_{x\to0} \frac{1}{x} \cdot \lim_{x\to0} (cosx^2-cosx)$
By the multiplication rule for limits
$\quad \quad \quad sin2 \cdot \lim_{x\to0} (\frac{1}{x} \cdot (cosx^2-cosx))$
Substituting  $x = 0$  for the latter expression
$\quad \quad \quad sin2 \cdot \lim_{x\to0} (\frac{1}{x} \cdot 0)$
$\quad \quad \quad sin2 \cdot \lim_{x\to0} (0)$
The second part is the limit of a constant, which is just the constant, therefore
$\quad \quad \quad sin2 \cdot 0$
Therefore, the entire second limit evaluates to $0$, and the answer to your question is the first limit; $-cos(2)$.
