<s>[INST] 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.
 [/INST] You can't, because you need to work with the interplay of the sine functions. Concretely, using Taylor approximations (and collecting the errors together), 
\begin{align}
\sin\dfrac{1}{x + 2} - 2 \sin\dfrac{1}{x + 1} + \sin\dfrac{1}{x} 
&=\frac1{x+2}-\frac1{6(x+2)^3}-2\left(\frac1{x+1}-\frac1{6(x+1)^3} \right)\\ \ \\
&\ \ \ \ \ \ \ \ \ \ \ \ +\frac1x-\frac1{6x^3}+o(\frac1{x^5})\\ \ \\
&=\frac2{x(x+1)(x+2)}-\frac1{6(x+2)^3}+\frac2{6(x+1)^3}\\ \ \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \  -\frac1{6x^3}+o(\frac1{x^5}).
\end{align}
Then
\begin{align}
x^3 \left ( \sin\frac{1}{x + 2} - 2 \sin\frac{1}{x + 1} + \sin\frac{1}{x} \right )
&=\frac2{(1+\tfrac2x)(1+\tfrac2x)}-\frac1{6(1+\tfrac2x)^3}\\ \ \\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ +\frac1{3(1+\tfrac1x)^3)}-\frac1{6x^3}+o(\tfrac1{x^2})\\ \ \\
&\xrightarrow[\vphantom{x_A}x\to\infty]{}2-\tfrac16+\tfrac13-\tfrac16=2.
\end{align}
\ 
</s>