<s>[INST] Find the value of $\lim _{x\to \infty }\left(x^2\ln \left(\frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}\right)\right)=$? $$\lim _{x\to \infty }\left(x^2\ln \left(\frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}\right)\right)=$$
I think it can be solved using L'Hospitals rule, but is there a simpler method?
 [/INST] A solution using Taylor expansions:
We will use the fact that, when $u\to 0$,
$$
\begin{align}
\sin u &= u -\frac{u^3}{6} + o(u^3) \\
\ln(1+u) &= u + o(u)
\end{align}
$$

Since $x\to\infty$, we have $\frac{1}{x}\to 0$, and we can apply the above with "$u=\frac{1}{x}$" (for the $\sin$) and "$u=\frac{-1}{6x^2} + o\left(\frac{1}{x^2}\right)$" (in the middle, for the $\ln$):
$$
\begin{align}
\ln \frac{\frac{1}{x}}{\sin \frac{1}{x}} 
&= - \ln \frac{\sin \frac{1}{x}}{\frac{1}{x}}
= - \ln \frac{\frac{1}{x} - \frac{1}{6x^3} + o\left(\frac{1}{x^3}\right)}{\frac{1}{x}} \\
&= - \ln\left( 1- \frac{1}{6x^2} + o\left(\frac{1}{x^2}\right) \right) \\
&= \frac{1}{6x^2} + o\left(\frac{1}{x^2}\right)  \\
\end{align}
$$
so that
\begin{align}
x^2\ln \frac{\frac{1}{x}}{\sin \frac{1}{x}} 
&= \frac{1}{6} + o\left(1\right)  \xrightarrow[x\to\infty]{} \frac{1}{6} \\
\end{align}
</s>