2 different limits for $x^x$ - Could they somehow be applied?

Question

So, I was just exploring the limit definition of e and seeing what I could create from it and after some time I landed at these 2 approximations for the power tower function - "$x^x$".

$ e(x^{x-1}+(x-1)^x) \approx x^x$

$ \frac{e(x-1)^{x+1}}{x-e} \approx x^x$

(I know they don't make much sense, but you can't separate out the x^x. Instead, they could used as large number approximations, or limits for e). I will explain how I got the first one - (as the second is just a re-arrangement fo the first using $x^{x-1}$ as the subject).

Also, for this I will be using the limit for the reciprocal of e --> $\lim_{x\to\infty} (1-\frac1x)^x = 1/e$

Proof for this is here

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$\lim_{x\to\infty} (1-\frac1x)^x = 1/e$

As 1/x approaches 0 as x goes to infinity, it can just be added on with changing the limit.

$\lim_{x\to\infty} (1-\frac1x)^x +\frac1x = 1/e$

$\lim_{x\to\infty} (1-\frac1x)^x +\frac{x^{x-1}}{x^x} = 1/e$

Then rewrite the inside of the bracket

$\lim_{x\to\infty} (\frac{x-1}{x})^x +\frac{x^{x-1}}{x^x} = 1/e$

$\lim_{x\to\infty} \frac{(x-1)^x}{x^x} +\frac{x^{x-1}}{x^x} = 1/e$

$\lim_{x\to\infty} \frac{{(x-1)^x} + {x^{x-1}}}{x^x} = 1/e$

Then just do some simple re-arrangement to get and discard the limit to get:

$ e(x^{x-1}+(x-1)^x) \approx x^x$

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I haven't seen this anywhere before, so I assume it's new.

My question is whether this could somehow be used to solve the integral of $x^x$ or define the function of $x^x$ in terms of an integral, just like the gamma function does for the factorial (after all, the factorial and power tower are similar functions)

I will also try using this to make an infinite continued fraction for $x^x$.

Answer 1

None of these identities make sense. The right-hand sides $x^x$ are meaningless as $x$ is not defined (or at best could be interpreted as $\infty^\infty$).

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Note that

$$\lim_{x\to\infty}\frac{e(x^{x-1}+(x-1)^x)}{x^x}=e\lim_{x\to\infty}\frac1x+e\lim_{x\to\infty}\left(\frac{x-1}x\right)^x=0+e\cdot e^{-1}$$ so that $$e(x^{x-1}+(x-1)^x)\sim x^x$$

but one may wonder if the first term has any relevance.

One can also wonder what use is to replace an expression by an approximation that is more complicated.

Answer 2

The fact of the matter that, just because two functions seem to have the same limit as $x\rightarrow\infty$ does not mean that the functions are necessarily close to each other. Because the function $x^x$ grows really quickly, and there is a lot of deviation you can have, while retaining the same limit.

For an example, the function ${\pi}(x)$ gives the number of primes less than or equal to $x$. One long-known fact is that the limit as $x\rightarrow\infty$ of $\pi(x)$ is the same as that of ${\rm li}(x):=x/\ln (x)$, in the sense that $\displaystyle\lim_{x\rightarrow\infty}\frac{\pi(x)}{{\rm li}(x)}=1$.

However, if you introduce a special metric to measure how good a certain approximation is, then you can see that ${\rm li}(x)-\frac 12{\rm li}(\sqrt x)$ is in fact a better approximation to $\pi(x)$. If you allow discontinuous functions, then in fact there is a whole range of other better estimates.

In other words, what I am saying is that, just because both functions seem to go to infinity in the same way does not mean that there is a special relationship between the two functions.