# Cleverly showing that $\lim_{x\to 0}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}=\frac{1}{6}$

$$\lim_{x\to 0}\frac{\textstyle x^{\textstyle(\sin x)^{\textstyle x}}-(\textstyle \sin x)^{\textstyle x^{\textstyle \sin x}}}{\textstyle x^3}=\frac{1}{6}$$

The limit is easy to get results, but how to rigorously prove it without using Taylor formula?

At first, I guess the numerator is equivalent to $$x-\sin x$$

And I also find the following several limits have the same results:

$$\lim_{x\to 0}\frac{\textstyle x^{\textstyle x^{\textstyle x}}-(\textstyle \sin x)^{\textstyle (\sin x)^{\textstyle \sin x}}}{\textstyle x^3}=\frac{1}{6}$$

$$\lim_{x\to 0}\frac{\textstyle x^{\textstyle x^{\textstyle \tan x}}-(\textstyle \sin x)^{\textstyle x^{\textstyle \tan x}}}{\textstyle x^3}=\frac{1}{6}$$

I guess if $$f(x)\sim g(x)\sim h(x)\sim O(x)^k$$ when $$x\to 0$$ , then $$\lim_{x\to 0}\textstyle f(x)^{\textstyle g(x)^{\textstyle h(x)}}=f(x)$$

• Can you use the rules of L'Hospital? Commented Apr 21, 2019 at 6:34
• @Dr. Sonnhard Graubner, It seems quite cumbersome to do that. The first derivative of a molecule=$\frac{x^{\sin ^x(x)} \sin ^x(x) (x \log (x) (x \cot (x)+\log (\sin (x)))+1)-x^{\sin (x)} \sin ^{x^{\sin (x)}}(x) (x \cot (x)+\log (\sin (x)) (\sin (x)+x \log (x) \cos (x)))}{x}$ Commented Apr 21, 2019 at 6:46
• In your last limit, you have $x\to0$ on the left and there is still $x$ on the right. Anyway, it's useless here: you don't want the limit of this, but the behaviour at order $3$. I would go with big-$O$ notation all along ("développement limité" in french, but I don't know the english translation). Commented Apr 21, 2019 at 7:25
• @Jean-ClaudeArbaut. I have the strange feeling that there is no translation beside Taylor expansion. Surprising, isn't it ? Cheers Commented Apr 21, 2019 at 8:32
• Perhaps what is required is a very general result along the last line, which goes into nested power limits and the likes. Let me do some preliminary investigations and see if results are available. Commented Jan 17, 2023 at 4:22

The expression under limit is of the form $$\frac {f(x) - g(x)} {x^3} =\frac{f(x) /x-g(x) /x} {x^2}$$ Now both $$f(x) /x, g(x) /x$$ tend to $$1$$ and hence we can replace them by their logarithms (see the lemma at the end). Thus the desired limit is equal to the limit of $$\frac{\log f(x) - \log g(x)} {x^2}$$ and this equals $$\frac {(\sin x) ^x\log x-x^{\sin x} \log\sin x} {x^2}$$ Now add and subtract $$x^{\sin x} \log x$$ in numerator to get $$\frac{(\sin x) ^x\log x-x^{\sin x} \log x}{x^2}-\frac{x^{\sin x} \log((\sin x) /x)} {x^2}$$ The last fraction tends to $$-1/6$$ so our job is done if we show that the first fraction above tends to $$0$$.

The first fraction can be written as $$\frac{(\sin x) ^x-x^{\sin x}} {x^2}\cdot \log x$$ Applying same technique again the limit of above fraction equals the limit of $$\frac{x\log \sin x-\sin x\log x} {x^2}\cdot\log x$$ Adding and subtracting $$x\log x$$ in numerator of the fraction we get $$x\log x\cdot\frac{\log((\sin x) /x)} {x^2}+\frac{x-\sin x} {x^3}\cdot x(\log x) ^2$$ Each of the terms above tends to $$0$$ and we are done.

In the above process we have used the following limits $$\lim_{x\to 0}\frac{\sin x} {x} =\lim_{x\to 0}\frac{e^x-1}{x}=1,\lim_{x\to 0^+}x^a(\log x) ^b=0,\forall a, b>0,\lim_{x\to 0}\frac{x-\sin x} {x^3}=\frac{1}{6}$$

The technique above can be used in a more general setting. Let us then suppose that each of the functions $$a(x), b(x), c(x), d(x), e(x), f(x)$$ is equivalent to $$x^n,n>0$$ as $$x\to 0^+$$ (ie $$a(x) /x^n\to 1, b(x) /x^n\to 1,\dots$$) and we need to evaluate the limit $$L$$ of fraction $$\frac{a(x) ^{b(x) ^{c(x)}} - d(x) ^{e(x) ^{f(x)}}} {x^m}, m>0$$ If the limits of the following fractions $$\frac {a(x) - d(x)} {x^m}, \frac {b(x) - e(x)} {x^m}, \frac {c(x) - f(x)} {x^m}$$ exist then the desired limit $$L$$ is equal to the limit of the first fraction above.

Lemma: Let $$f, g, h$$ be real valued functions defined in a deleted neighborhood of $$a$$ such that $$\lim_{x\to a} f(x) =\lim_{x\to a} g(x) =L>0$$ Then the limiting behavior of $$(f(x) - g(x))^{\pm 1} h(x)$$ is same as that of $$(L(\log f(x) - \log g(x))) ^{\pm 1}h(x)$$ as $$x\to a$$.

We can write $$f(x) - g(x) = g(x) \cdot\left(\frac{f(x)} {g(x)} - 1\right)$$ Next note that $$f(x) /g(x) \to 1$$ and if $$t=\log(f(x) /g(x))$$ then $$t\to 0$$ and the above expression can be written as $$g(x) \cdot\frac{e^t-1}{t}\cdot t$$ and thus we can replace the above with $$L\cdot t$$ ie $$L(\log f(x) - \log g(x))$$. The given conditions ensure that $$f, g$$ are positive so their logs make sense.

Further note that the expression $$g(x) ((e^t-1)/t)$$ occurs in a multiplicative manner (ie like a factor) in the overall expression $$(f(x) - g(x)) ^{\pm 1}h(x)$$ and has a non-zero limit $$L$$ and thus we can safely replace it by $$L$$ without worrying about the limiting behavior of remaining part of the expression. For more details about replacing sub-expressions with their limits one can refer to this answer as well as its more formal version discussed in this thread.

The solution in this answer uses this lemma with $$L=1$$.

• This is a nice answer; general and specific. I'll probably be awarding it the bounty. Commented Jan 18, 2023 at 18:40
• @Integrand: I updated the answer with a more general result. Commented Jan 19, 2023 at 5:10
• I will write $F(x)$ for $f(x)/x$. There is a claim at the beginning regarding $f(x)/x$ and $g(x)/x$, for this claim we can take $g=1$, and then it comes to $F(x)\to 1$ for $x\to 0$ then there exist in the same time and are equal the limits for $x\to 0$, $x>0$:$$\lim\frac{F(x)-1}{x^2}=\lim\frac{\log F(x)-\log 1}{x^2}\ .$$How do we show this (without assuming a Taylor expansion $F(x)=1+ax+bx^2+\dots$, so that $\log F(x)=ax +\left(b-\frac 12a^2\right)x^2+\dots$ and for $a\ne 0$ both sides are infinite, for $a=0$...) ? Commented Jan 21, 2023 at 19:51
• @dan_fulea: I have updated my answer with a lemma at the end which provides more details. Commented Jan 22, 2023 at 1:10

Note the limit is only defined as $$x\to 0^+$$, so I'll answer as such. Not sure how 'clever' this is either and the solution is a little fast and loose, but oh well.

1. Using LHR or other methods we have $$\lim_{x\to0^+} f_1(x)^{f_2(x)} = 1,$$for $$f_1,f_2\in \{\sin,\operatorname{id}\}$$. This means we can interchange some of the exponents with mild impunity. In particular, I claim $$\lim_{x\to0^+} \frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3} =\lim_{x\to0^+} \frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x^3}$$
2. Introduce $$x-\sin x$$: $$=\lim_{x\to0^+} \frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x-\sin x}\cdot \frac{x-\sin x}{x^3}$$We will break into two limits and justify this by showing the first equals $$1$$; the second clearly equals $$1/6$$ by LHR. Then we are interested in $$\lim_{x\to0^+} \frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x-\sin x}$$
3. Let $$q:[0,1]\to\mathbb{R}$$, $$q(z)=z^{z^z}$$ and $$q(0)=0$$. By the MVT, for some $$y$$ between $$\sin(x)$$ and $$x$$ we have $$\frac{x^{x^x}-(\sin x)^{(\sin x)^{\sin x}}}{x-\sin x} = q'(y)$$Since $$q'$$ is continuous, it suffices to show $$\lim_{y\to 0^+}q'(y) = 1$$.
4. We have $$q'(y) = y^{y^y+y-1} \left(y \log ^2 y+y \log y+1\right)$$The second term is easily seen to approach $$1$$ in the limit; for example, put $$y=e^w$$ or play with LHR. The first term isn't as apparent, but using the bound $$1>y^y>1+y\log y$$ on $$(0,1)$$ reduces the problem to evaluating $$\lim_{y\to 0^+}y^{y+y\log y},$$ which is just $$1$$ (exercise). One could also show it directly.
• Your first claim does no hold (in general). You cannot substitute an expression in a limit calculation by another one with the same limit and expect the original limit to be preserved. For instance, $$\lim_{x\to 0}\frac{x-x}{x} \ne \lim_{x\to 0}\frac{x^2-x}{x},$$ even though $\lim_{x\to 0} x = \lim_{x\to 0} x^2$. Commented Jan 17, 2023 at 15:48
• I didn't claim it held in general; I claimed, admittedly without proof, that it holds in this instance. Perhaps I will supply the missing details, but the rest of it holds with that assumption. Commented Jan 17, 2023 at 15:57
• Not quite. You used the trick more than once, so the rest of it holds under $\color{red}{\rm those\ assumptions}$ Commented Jan 19, 2023 at 12:04

This is actually a long comment and doesn't fully answer the question. But I hope it turns out to be helpful somehow.

Regarding the general case, assume that $$l(x)=f(x)^{g(x)^{h(x)}}$$ and in a right neighborhood of $$0$$: $$f(x)\sim g(x)\sim h(x)=o(x^k), \qquad k\ge 1$$ Then $$l'(x)=g(x)^{h(x)}\;l(x)\left(\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}h(x)\log f(x)+h'(x)\log g(x) \log f(x)\right)$$ Now both $$\frac{f'}f$$ and $$\frac{g'}g$$ are $$o(x^{-1})$$ and $$\log f\sim \log g=o(\log x)$$. Therefore: $$\frac{l'(x)}{l(x)}=g(x)^{h(x)}\left(o(x^{-1})+o(x^{k-1}\log x)+o(x^{k-1}\log^2 x)\right)$$ From there, I think it would be easier to validate your hypothesis.

• This is a good idea but it's not immediately clear to me that the last line is sufficient, though maybe I'm just not seeing it. Commented Jan 18, 2023 at 18:41

Too long for a comment :

Some tought around zero or $$x\in(0,1/10)$$ we have :

$$\left(\sin\left(x\right)\right)\left(\sin\left(x\right)-1\right)\left(\left(\sin\left(x\right)\right)^{x}-2\right)-\left(\sin\left(x\right)^{\left(x^{\sin\left(x\right)}\right)}\right)\geq \left(x\left(x-1\right)\left(x^{\sin\left(x\right)}-2\right)\right)-\left(x^{\left(\sin\left(x\right)^{x}\right)}\right)$$

So using this inequality we have as limit :

$$\lim_{x\to 0^+}=\frac{\left(x\left(x-1\right)\left(x^{\sin\left(x\right)}-2\right)\right)-\left(\sin\left(x\right)\right)\left(\sin\left(x\right)-1\right)\left(\left(\sin\left(x\right)\right)^{x}-2\right)}{x^{3}}=\lim_{x\to 0^+}\frac{\left(x\left(x-1\right)\left((\sin(x))^{x}-2\right)\right)-\left(\sin\left(x\right)\right)\left(\sin\left(x\right)-1\right)\left(\left(\sin\left(x\right)\right)^{x}-2\right)}{x^{3}}=1/6$$

As for $$x\geq a$$ such that $$a,x\in(0,1)$$ :

$$a\left(a-1\right)\left(a^{x}-2\right)-\left(a^{x^{a}}\right)=g(x),f(x)=x\left(x-1\right)\left(x^{a}-2\right)-\left(x^{a^{x}}\right)$$

$$g(x)$$ is increasing and $$f(x)$$ is decreasing .

We have in fact for $$x\ge y$$ such that $$x,y\in(0,1)$$ :

$$y\left(y-1\right)\left(y^{x}-2\right)-y^{x^{y}}-x\left(x-1\right)\left(x^{y}-2\right)+x^{y^{x}}\geq 0$$

As $$\sin(x)\leq x$$ the desired inequality follow .

• @Integrand What do you think about ? Commented Jan 20, 2023 at 17:10
• Interesting idea. Where did the first inequality come from? Commented Jan 20, 2023 at 18:47
• @Integrand It's a fail attempt to use Bernoulli's inequality . Commented Jan 21, 2023 at 10:52

The process of evaluating this limit consists of three steps:
Step 1 - Show that:
$$lim_{x\to 0}(sin(x))^{x} = 1$$
and
$$lim_{x\to 0}(x)^{sin(x)} = 1$$
Step 2- Simplify the limit using the results from Step 1
Step 3 - Apply L'Hopital's rule three times to the limit which was simplified in Step 2

Step 1
Evaluate the first limit:
$$lim_{x\to 0}(sin(x))^{x}=lim_{x\to+0}e^{ln(sin(x)^{x})}=e^{lim_{x\to+0}(x\cdot ln(sin(x)))}=e^{lim_{x\to+0}\frac{ln(sin(x))}{\frac{1}{x}}}$$
Now, apply L'Hopital's rule to the limit in the exponent:
$$e^{lim_{x\to+0}\frac{ln(sin(x))'}{(\frac{1}{x})'}}=e^{lim_{x\to+0}\frac{\frac{cosx}{sinx}}{-(\frac{1}{x^{2}})}}=e^{-lim_{x\to+0}\frac{cosx\cdot x^{2}}{sinx}}=e^{-lim_{x\to+0}\frac{cosx\cdot x}{\frac{sinx}{x}}}=e^{0}=1$$
The second limit can be converted to the first one using the first remarkable limit:
$$lim_{x\to0}(x^{sin(x)})=lim_{x\to0}((x\cdot 1)^{(sin(x)\cdot 1)})=lim_{x\to0}((x\cdot \frac{sin(x)}{x})^{(sin(x)\cdot \frac{x}{sin(x)})}=lim_{x\to0}((sin(x))^{x})=lim_{x\to+0}((sin(x))^{x})=1$$

Step 2
Simplify the original limit using the results of Step 1:
$$lim_{x\to0}\frac{x^{(sinx)^{x}}-(sinx)^{x^{sinx}}}{x^{3}}=lim_{x\to0}\frac{x^{lim_{x\to0}(sinx)^{x}}-(sinx)^{lim_{x\to0}x^{sinx}}}{x^{3}}=lim_{x\to0}\frac{x-sinx}{x^{3}}$$ Step 3
Apply L'Hopital's rule three times:

$$lim_{x\to0}\frac{(x-sinx)'''}{(x^{3})'''} = lim_{x\to0}\frac{(1-cosx)''}{(3x^{2})''}=lim_{x\to0}\frac{(sinx)'}{(6x)'}=lim_{x\to0}\frac{cosx}{6}=\frac{1}{6}$$

• This is essentially what all the other answers do, your solution does not bring anything new into the discussion. Commented Jan 21, 2023 at 8:47
• You can't replace a sub-expression by its limit during evaluation of limit of the expression (second limit in step 1 as well complete step 2). This is a fundamental mistake. - 1 Also please fix your mathjax. Commented Jan 21, 2023 at 9:06
• Thank you Alex for your prompt comment. Can you please clarify to me why none of the previous answers have been accepted yet? Commented Jan 21, 2023 at 9:20