Evaluate $$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right].$$


First, we may obtain $$\lim_{x \to 0+}\left[\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}\right]=\lim_{x \to 0+}\frac{6e^{\sin x\ln x}-6e^{x\ln\sin x}+x^3\ln x}{6x^3}.$$ Here, you can apply L'Hôpital's rule, but it's too complicated. Moreover, you can also apply Taylor's formula, for example $$e^{\sin x\ln x}=1+\sin x\ln x+\frac{1}{2}(\sin x\ln x)^2+\cdots,\\e^{x\ln\sin x}=1+x\ln\sin x+\frac{1}{2}(x\ln\sin x)^2+\cdots,$$ but you cannot cancel the terms, thus you cannot avoid differentiating, either. Is there any elegant solution?

P.S. Please don't suspect the existence of the limit. The result equals $\dfrac{1}{6}.$


The key point is that $x\log \sin x \to 0$ and $\sin x \log x \to 0$ then by Taylor's series we have

  • $x^{\sin x}=e^{\sin x \log x}=1+x\log x+\frac12x^2\log^2 x+\frac16x^3\log x(\log^2 x -1)+O(x^4\log^2 x)$
  • $(\sin x)^{x}=e^{x \log (\sin x)}=1+x\log x+\frac12x^2\log^2 x+\frac16x^3(\log^3 x -1)+O(x^4\log x)$


$$\frac{x^{\sin x}-(\sin x)^{x}}{x^3}+\frac{\ln x}{6}=\frac{\frac16x^3\log^3 x-\frac16x^3\log x-\frac16x^3\log^3 x +\frac16x^3+O(x^4\log x)}{x^3}+\frac{\ln x}{6}=$$

$$=\frac16+O(x\log x) \to \frac16$$

To see how obtain the Taylor's expansion, let consider the first one, then since

  • $\sin x =x-\frac16 x^3+O(x^5) \implies \sin x \log x=x\log x-\frac16 x^3\log x+O(x^5\log x)$
  • $e^t = 1+t+\frac12 t^2+\frac16t^3+O(t^4)$

we obtain that

$$x^{\sin x}=e^{\sin x \log x} =1+\left(x\log x-\frac16 x^3\log x+O(x^5\log x)\right)+\frac12\left(x\log x-\frac16 x^3\log x+O(x^5\log x)\right)^2+\frac16\left(x\log x-\frac16 x^3\log x+O(x^5\log x)\right)^3+O(x^5\log^4 x)=$$

$$=1+x\log x-\frac16 x^3\log x+\frac12x^2\log^2x-\frac16x^4\log^2x+\frac16x^3\log^3x+O(x^4\log^2x)=$$

$$=1+x\log x+\frac12x^2\log^2x+\frac16x^3\log x(\log^2x-1)+O(x^4\log^2x)$$

and for the second one since

  • $\log (1+t)= t-\frac12t^2+\frac13t^3+O(t^4)$
  • $\sin x =x-\frac16 x^3+O(x^5)\implies \frac{\sin x}x=1-\frac16 x^2+O(x^4)$
  • $\log \sin x=\log x+\log \frac{\sin x}x=\log x+\log \left(1-\frac16 x^2+O(x^4)\right)=\log x-\frac16 x^2+O(x^4)$
  • $x\log \sin x=x\log x-\frac16 x^3+O(x^5)$

we obtain that

$$(\sin x)^x=e^{x\log \sin x}=1+\left(x\log x-\frac16 x^3+O(x^5)\right)+\frac12\left(x\log x-\frac16 x^3+O(x^5)\right)^2+\frac16\left(x\log x-\frac16 x^3+O(x^5)\right)^3+O(x^4\log^4x)$$

$$=1+x\log x-\frac16 x^3+\frac12x^2\log^2x-\frac16x^4\log x+\frac16x^3\log^3 x+O(x^4\log x)=$$

$$=1+x\log x+\frac12x^2\log^2x+\frac16x^3(\log^3 x-1)+O(x^4\log x)$$

  • $\begingroup$ How to derive Taylor‘s Formula in such form? $\endgroup$ – mengdie1982 Nov 10 '18 at 11:34
  • $\begingroup$ $$e^{\sin x\ln x}=1+\sin x\ln x+\frac{1}{2}(\sin x\ln x)^2+\cdots,$$ $\endgroup$ – mengdie1982 Nov 10 '18 at 11:38
  • $\begingroup$ @mengdie1982 Starting form $x^{\sin x}=e^{\sin x \log x}$ and then expanding then $\sin x \log x$ and then using that $x^a \log x \to 0$ for any $a>0$ to expand the exponential. It is not a short way but it is not difficult, $\endgroup$ – gimusi Nov 10 '18 at 11:38
  • $\begingroup$ Put $\sin x=1+\cdots$ into it? $\endgroup$ – mengdie1982 Nov 10 '18 at 11:39
  • $\begingroup$ @mengdie1982 Yes that's also good but I would prefer at frst expand sin x log x and then the exponential. $\endgroup$ – gimusi Nov 10 '18 at 11:39

The key here is obtaining the expansions for $x^{\sin x} $ and $\sin^xx$ but it is simpler to deal with their logarithms. Consider \begin{align} f(x) &=\sin x \log x-x\log\sin x\notag\\ &=(x\log x)\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots \right)-x\log x-x\log\left(1-\frac{x^2}{6}+\frac{x^4}{120}-\dots\right)\notag\\ &=\notag\\ &=-\frac{x^3\log x} {6}+\frac{x^3}{6}+o(x^3)\notag \end{align} And then the expression under limit is $$\sin^xx\cdot\frac{e^{f(x)} - 1}{x^3}+\frac{\log x} {6}$$ which can be further written as $$\sin^xx \left(\frac{e^{f(x)} - 1}{x^3}+\frac{\log x} {6}\right) +\frac{\log x} {6}\cdot(1-\sin^xx) $$ From the expansion of $f(x) $ it is clear that $f(x) =o(x^2)$ and hence $$\frac{e^{f(x)} - 1}{x^3}=-\frac{\log x} {6}+\frac{1}{6}+o(1)$$ The desired limit is $1/6$ if we can prove that $\sin^xx \to 1$ and $(\log x) (1-\sin^xx) \to 0$ and this is not difficult to prove.

We just need to note that $$(\log x) \cdot(1-\sin^xx)=-\frac{\exp(x\log\sin x) - 1}{x\log\sin x} \cdot x\log \sin x\cdot\log x$$ and the fraction above tends to $1$ so that the limit of above expression is equal to the limit of $$-x\log x\left(\log x+\log\frac{\sin x} {x} \right) $$ which is same as $$-x(\log x) ^2-(x\log x) \log\frac{\sin x} {x} $$ and the above clearly tends to $0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.