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.

Let $t=\frac1x$. Then,
$$\lim_{x \to \infty} x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
=\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t \right )$$
Use $\frac 1{1+a} = 1-a+a^2+O(a^3)$ to expand,
$$\sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t } + \sin t$$
$$=\sin(t-2t^2+4t^3)+\sin t  - 2 \sin(t-t^2+t^3)+O(t^4)$$
$$=2\sin(t-t^2+2t^3)\cos t^2 - 2 \sin(t-t^2+t^3)+O(t^4)$$
$$=2[\sin(t-t^2+2t^3) - \sin(t-t^2+t^3)]+O(t^4)$$
$$=4\cos t\sin\frac{t^3}2+O(t^4)= 4\cdot 1\cdot \frac{t^3}2+O(t^4)=2t^2+O(t^4)$$
where $\cos t^2 = 1 + O(t^4)$ is applied. Thus,
$$\lim_{t \to 0} \frac1{t^3} \left ( \sin\frac{t}{1 + 2t} - 2 \sin\frac{t}{1+t} + \sin t \right )=\lim_{t \to 0}  \frac{2t^3+O(t^4)} {t^3}=2$$
