From the limit of a function to its value and its derivatives at a point.

Let $f$ be a infinitely differentiable function such tha

$$\lim_{x\to 0}\frac{2f(x)+2x^2-6\cos x}{3x^2}=0.\tag 1$$

I have to calculate the following quantity: $$f(0)+f'(0)+f^{''}(0).$$

My solution.

Since $(1)$ hold then $$f(x)\to -x^2+3\cos x\quad\text{for}\; x\to 0$$ and therefore for $x\to 0$ we have that $f(x)\to 3$.

Now, $f'(x)\to -2x-3\sin x$ when $x\to 0$ and therefore $$f'(x)\to 0\quad\text{for}\; x\to0.$$ In the same way $f^{''}(x)\to -2-3\cos x$ for $x\to 0$ and then $f^{''}(x)\to -5$.

Combining all the obtained results, we have that $$f(0)+f^{'}(0)+f^{''}(0)=3+0-5=-2.$$

QuestionI'm not sure if the procedure is correct, it seems to me that I'm using convergence of functions. Is this type of procedure correct? Otherwise, how should I proceed? Thanks in advance!