Limit of x² ln((1/x) / sin(1/x))
Evaluate the limit: $$ L = \lim_{x \to \infty} x^2\ln\left(\frac{\frac1x}{\sin\frac1x}\right). $$
Explanation of Method
Substituting $u=\frac1x$ turns the limit at infinity into a limit as $u\to0$, of exactly the same shape as the well-known small-angle limit $\ln\left(\frac{\sin u}u\right)\sim-\frac{u^2}6$. Once the substitution is made, the logarithm's argument is recognized as the reciprocal of $\frac{\sin u}u$, so the logarithm just picks up an overall minus sign, and the whole problem reduces to a limit already familiar from other small-angle expansions.
- Substitute $u=\frac1x$, converting the limit to $u\to0$.
- Rewrite the logarithm's argument as the reciprocal of $\frac{\sin u}{u}$.
- Expand $\ln\left(\frac{\sin u}u\right)$ using the Taylor series for $\sin u$.
- Combine and take the limit.
Step 1 · Substitute $u=1/x$
Let $u=\dfrac1x$, so $x=\dfrac1u$ and $x^2=\dfrac1{u^2}$. As $x\to\infty$, $u\to0$. The limit becomes:
$$ L = \lim_{u\to0} \frac{1}{u^2}\ln\left(\frac{u}{\sin u}\right). $$
Step 2 · Rewrite as a reciprocal
Since $\dfrac{u}{\sin u} = \left(\dfrac{\sin u}{u}\right)^{-1}$, and $\ln(z^{-1})=-\ln z$:
$$ \ln\left(\frac{u}{\sin u}\right) = -\ln\left(\frac{\sin u}{u}\right). $$
So:
$$ L = -\lim_{u\to0}\frac{1}{u^2}\ln\left(\frac{\sin u}{u}\right). $$
Step 3 · Expand $\ln\left(\dfrac{\sin u}u\right)$
The Taylor series for $\sin u$ near $0$ is $\sin u = u-\dfrac{u^3}6+O(u^5)$, so:
$$ \frac{\sin u}{u} = 1-\frac{u^2}{6}+O(u^4). $$
Using $\ln(1+z)\approx z$ for small $z$, with $z=-\dfrac{u^2}6+O(u^4)$:
$$ \ln\left(\frac{\sin u}{u}\right) \approx -\frac{u^2}{6}. $$
Step 4 · Combine and take the limit
Substituting into the expression for $L$:
$$ L = -\lim_{u\to0}\frac{1}{u^2}\left(-\frac{u^2}{6}\right) = -\left(-\frac16\right) = \frac16. $$
Summary Table
| Quantity | In terms of $u=1/x$ | Behavior as $u\to0$ |
|---|---|---|
| $\dfrac{\sin u}{u}$ | $1-\dfrac{u^2}{6}+O(u^4)$ | $\to1$ |
| $\ln\left(\dfrac{\sin u}u\right)$ | $-\dfrac{u^2}6+O(u^4)$ | $\to0$ |
| $L$ | $-\dfrac1{u^2}\ln\left(\dfrac{\sin u}u\right)$ | $\to\dfrac16$ |
Additional Notes
- This problem is really the same computation as the small-angle limit $\lim_{u\to0}\frac{1}{u^2}\ln\left(\frac{\sin u}{u}\right)=-\frac16$ seen elsewhere, in disguise: substituting $u=\frac1x$ converts "at infinity" behavior into "near $0$" behavior, and flipping the fraction inside the log just introduces an overall sign.
- Recognizing a substitution that turns an unfamiliar limit into a previously solved one is a powerful general strategy — always check whether $u=\frac1x$ (or another substitution) reduces a new-looking problem to a standard form.
- Only the leading $u^2$ term of the expansion of $\frac{\sin u}u$ mattered, since it is divided by $u^2$ before taking the limit — this same $-\frac16$ coefficient shows up any time $\sin u/u$ appears logarithmically divided by $u^2$.
Result
The value of the limit is: