Find the value of $\lim _{x\to \infty }\left(x^2\ln \left(\frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}\right)\right)=$?

Question

$$\lim _{x\to \infty }\left(x^2\ln \left(\frac{\frac{1}{x}}{\sin \left(\frac{1}{x}\right)}\right)\right)=$$

I think it can be solved using L'Hospitals rule, but is there a simpler method?

Answer 1

A solution using Taylor expansions:

We will use the fact that, when $u\to 0$, $$ \begin{align} \sin u &= u -\frac{u^3}{6} + o(u^3) \\ \ln(1+u) &= u + o(u) \end{align} $$

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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}