# Finding $\lim_{x \to \infty} \left[ {x^{x+1} \over (x+1)^x} - { (x-1)^x\over x^{x-1}}\right]$

The limit is $$\lim_{x \to \infty} \left[ {x^{x+1} \over (x+1)^x} - { (x-1)^x\over x^{x-1}}\right]$$

Experimentally, this limit appears to converge to ${1 \over e}$, but I can't figure out how to solve it.

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Mathematica confirms the limit is 1/e; as for figuring out how this result was arrived at, just to give you a hint, this is an ∞-∞ type indeterminate form. Manipulate it into something where L'Hôpital can apply (you may also have to invoke logarithmic differentiation at some point). Good luck! – J. M. Aug 14 '10 at 0:30
I did manipulate it and use L'Hôpital's rule, but then it seemed to explode into further complexity, so I gave up. If this is the proper way to solve the limit, then I'll try to muck through all the details. – dln385 Aug 14 '10 at 0:41
You will have to take logarithms at some point and come up with an expression whose limit is -1. But since this is the limit for the logarithm, the limit for the original expression should be $\exp(-1)$. – J. M. Aug 14 '10 at 1:01

A fairly mechanical approach is to write the limit as $$\lim_{x\to\infty}(f(x)-f(x-1))$$ where $$f(x)=\frac{x^{x+1}}{(x+1)^x}.$$ Then $$\log f(x)=(x+1)\log x - x\log(x+1) = \log x - x \log(1+1/x)$$ and so $$\log f(x)= \log x - 1 + 1/(2x) + O(x^{-2})$$ as $x\to\infty$ (using the Maclaurin series for $\log(1+t)$). Therefore $$f(x) = (x/e)(1+1/(2x)+O(x^{-2}))=x/e+1/(2e)+O(x^{-1})$$ and so $$f(x-1) =(x-1)/e+1/(2e) +O((x-1)^{-1}).$$ Subtracting, $$f(x)-f(x-1)= 1/e+O(x^{-1})).$$

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Making the change of variables $\; z = 1/x \;$ and collecting together like exponents yields

$\quad\quad\quad\quad \displaystyle \lim_{z\to 0^+} {\frac{(1+z)^{-1/z} - (1-z)^{1/z}}{z}}\; = \;\lim_{z\to 0^+} {\frac{f(-z)-f(z)}{z}}$

Applying the well-known $\rm exp$ and $\rm log$ taylor series to $f(z) = (1-z)^{1/z}\; = \; e^{{\rm log}(1-z)/z}$

$$f(z) = e^{-1-\frac{z}{2}+\;\cdots} = e^{-1} (1 + (-\frac{z}{2} +\;\cdots) + (-\frac{z}{2} +\cdots)^2 + \;\cdots) = \frac{1}{e} - \frac{z}{2e} + O(z^2)$$

Therefore $\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle \frac{f(-z)-f(z)}{z}\; = \;\frac{1}{e} + O(z)\quad\quad\quad$ as confirmed by Macsyma:

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+1 very cool :) – Jonathan Fischoff Aug 14 '10 at 1:50
+1. You're gone but we are looking to see you again on. – S. Snape Nov 7 '13 at 6:12

HINT You can write the difference as $x \left( y^{-x} - z^x \right)=x\left(1-(yz)^x\right)/y^z$, where $y$ and $z$ are rational functions of $x$ such that the limiting value of $y^x$ (and $z^x$) is well-known, finite, and non-zero. Now focus on finding the limit of $x\left(1-(yz)^x\right)$.

(Restricting $x$ to integers and invoking the Binomial Theorem helps, but that only shows what the limit --if it exists-- would have to be, leaving open the possibility that the limit might not exist when considering non-integer $x$.)

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Well, the binomial theorem still works for noninteger exponents (we now have an infinite series because none of the binomial coefficients vanish), though it no longer applies if the variable of interest is both in the base and the exponent. – J. M. Aug 14 '10 at 14:45