Find $\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )$ I have the following limit to find:
$$\lim\limits_{x \to \infty} x^3 \bigg ( \sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} \bigg )$$
What approah should I use? Since it's an $\infty \cdot 0$ type indeterminate I thought about writing $x^3$ as $\dfrac{1}{\frac{1}{x^3}}$ so I would have the indeterminate form $\dfrac{0}{0}$, but after applying L'Hospital I didn't really get anywhere.

Using first approximation for $\sin x\approx x$ for $x$ near $0$, the limit can be rewritten without change of variables as $$\begin{aligned} &\lim_{x\to \infty}x^3\left(\frac{1}{x+2}-\frac{2}{x+1}+\frac{1}{x}\right)\\ = &\lim_{x\to \infty}x^3\left[\left(\frac{1}{x}-\frac{1}{x+1}\right)-\left(\frac{1}{x+1}-\frac{1}{x+2}\right)\right]\\=&\lim_{x\to \infty}x^3\left[\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}\right]\\=&\lim_{x\to \infty}\frac{2x^3}{x(x+1)(x+2)}\to 2\end{aligned}$$
