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

Problem

Evaluate

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

Attempt

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 $\dfrac{1}{6}.$

Answer

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

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


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

then

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

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

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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

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

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

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