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

Question

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

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 $$\lim_{x\to 0}\textstyle f(x)^{\textstyle g(x)^{\textstyle h(x)}}=f(x)$$

Answer 1

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 $$\lim_{x\to 0}\frac{\sin x} {x} =\lim_{x\to 0}\frac{e^x-1}{x}=1,\lim_{x\to 0^+}x^a(\log x) ^b=0,\forall a, b>0,\lim_{x\to 0}\frac{x-\sin x} {x^3}=\frac{1}{6}$$

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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.

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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 $L=1$.

Answer 2

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.

Answer 3

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.

Answer 4

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 .

Answer 5

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