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Okay, this is the last limit I have to solve but it's not that easy ;)

$$\lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$$

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Yeah, i did the smae but the reasoning is not sounding really legit :D – Radoslav Georgiev Aug 28 '11 at 16:14
With these limit questions why not try out some actual values to get a sense of what's happening? – Mark Bennet Aug 28 '11 at 16:48
@Ilmari, actually your "indication" is absolutely unsound (and I find most welcome Radoslav's instinctive scepticism here). You could try to test the example $x^{1+1/\log(x)}−x$. You would probably predict that the limit is $0$, and this would be wrong. – Did Aug 28 '11 at 16:51
@Ilmari, yes the respective order of our comments is surprising but this phenomenon escapes me. – Did Aug 28 '11 at 17:16
@Ilmari, $(x+c)^{1+1/x}$ does not approach $x+c$, at least not in the sense that their difference would go to zero. – Did Aug 28 '11 at 17:20
up vote 15 down vote accepted

One needs to be careful because this is a difference of two functions and each of them converges to infinity, hence the behaviour of their difference can be almost anything.

Write the function of interest as $f(x)=xx^{1/x}g(x)$ with $$ g(x)=(1+3/x)^{1+1/x}-x^{1/(x+3)-1/x}=\mathrm{e}^{a(x)}-\mathrm{e}^{b(x)}, $$ with $$ a(x)=(1+1/x)\log(1+3/x)=3/x+o(1/x), $$ and $$ b(x)=-\frac{3\log(x)}{x(x+3)}=o(1/x). $$ This yields $\mathrm{e}^{a(x)}=1+3/x+o(1/x)$ and $\mathrm{e}^{b(x)}=1+o(1/x)$ hence $g(x)=3/x+o(1/x)$.

Likewise $x^{1/x}=\mathrm{e}^{\log(x)/x}=1+o(1)$ hence $$ f(x)=xx^{1/x}g(x)=x(1+o(1))(3/x+o(1/x))=3+o(1). $$ That is, $f$ does converge and its limit is $3$.

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Wow :) Thanks for the solution, looks great. – Radoslav Georgiev Aug 29 '11 at 0:11
@Radoslav, you are welcome. – Did Sep 1 '11 at 22:15

I know the following method is not valid for the purpose of this exercise, but I would like to post it. I computed in the Computer Algebra System included in SWP the following power series expansions:

$$\begin{eqnarray*} (t^{-1}+3)^{1+t} &=&t^{-1}+\left( 3-\ln t\right) +\left( -\frac{3}{2}+\frac{1 }{2}\left( 3-\ln t\right) ^{2}\right) t+O\left( t^{2}\right) \\ t^{-1-1/(t^{-1}+3)} &=&t^{-1}+\left( -\ln t\right) +\left( 3\ln t+\frac{1}{2} \ln ^{2}t\right) t+O\left( t^{2}\right) \end{eqnarray*}$$


$$(t^{-1}+3)^{1+t}-t^{-1-1/(t^{-1}+3)}=3+\left( -\frac{3}{2}+\frac{1}{2}\left( 3-\ln t\right) ^{2}-3\ln t-\frac{1}{2}\ln ^{2}t\right) t+O\left( t^{2}\right).$$

Then I generated the asymptotic power series of $(x+3)^{1+1/x}-x^{1+1/(x+3)}$ by the change of variables $x=1/t$

$$\begin{eqnarray*} f(x)&=&(x+3)^{1+1/x}-x^{1+1/(x+3)} \\ &=&3+\left( -\frac{3}{2}+\frac{1}{2}\left( 3-\ln \frac{1}{x}\right) ^{2}-3\ln \frac{1}{x}-\frac{1}{2}\ln ^{2}\frac{1}{x} \right) \frac{1}{x}+O\left( \frac{1}{x^{2}}\right) \\ &=&3+\left( -\frac{3}{2x}+\frac{1}{2x}\left( 3+\ln x\right) ^{2}+\frac{3\ln x }{x}+\frac{\ln ^{2}x}{2x}\right) +O\left( \frac{1}{x^{2}}\right) \\ &=&3+\frac{3}{x}+6\frac{\ln x}{x}+\frac{\ln ^{2}x}{x}+O\left( \frac{1}{x^{2}} \right) \\ &\sim &3, \end{eqnarray*}$$

and confirmed with the following variant $$\begin{eqnarray*} f(x)&=&(x+3)^{1+1/x}-x^{1+1/(x+3)} \\ &=&\frac{\left( x+3\right) ^{1+1/x}x^{-1-1/(x+3)}-1}{x^{-1-1/(x+3)}} \\ &\sim &\frac{1+\frac{3}{x}+\left( -3\ln \frac{1}{x}+3\right) \frac{1}{x^{2}} +O\left( \frac{1}{x^{3}}\right) -1}{\frac{1}{x}+\left( \ln \frac{1}{x} \right) \frac{1}{x^{2}}+O\left( \frac{1}{x^{3}}\right) } \\ &\sim &3\frac{x+\ln x+1}{x-\ln x} \\ &\sim &3. \end{eqnarray*}$$

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$(x+3)e^{((\ln(x+3))/x)}-xe^{((\ln x)/(x+3))}= 3+{(((x+3)\ln(x+3))/x)-((x\ln x)/(x+3))}+o(1)= 3+o(1)$, where the first equality is obtained by Maclaurin series of the exponential function and the last equality is obtained by the trick $\ln(x+3)= \ln x+\ln(1+(3/x))$.

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For constants a, b, consider the asymptotics of $(x+a)^{1+1/(x+b)}$ as $x\to\infty$. This is equal to $(x+a)e^{\log(x+a)/(x+b)}$.

Now since $\log(x+a)/(x+b)$ tends to 0, we know that $e^{\log(x+a)/(x+b)}=1+\log(x+a)/(x+b)+\ldots$ where the $\ldots$ term tends to 0 faster than $\log(x+a)^2/(x+b)^2$.

Then $(x+a)^{1+1/(x+b)}=(x+a)+(x+a)\log(x+a)/(x+b)+(x+a)\ldots$, and since $(x+a)\log(x+a)^2/(x+b)^2$ tends to 0, so does $(x+a)\ldots$.

Therefore $(x+a)^{1+1/(x+b)}-(x+a)-(x+a)\log(x+a)/(x+b)\to0$ as $x\to\infty$.

Note that $(x+a)\log(x+a)/(x+b)-\log(x+a)=(a-b)\log(x+a)/(x+b)\to0$ as $x\to\infty$.

Note also that $\log(x+a)-\log x=\log(1+a/x)\to\log 1=0$ as $x\to\infty$.

Adding these three limits together shows that $(x+a)^{1+1/(x+b)}-(x+a)-\log x\to 0$, which we can rewrite as $(x+a)^{1+1/(x+b)}-(x+\log x)\to a$.

Applying with $a=3$ and $b=0$ gives that $(x+3)^{1+1/x}-(x+\log x)\to 3$. Applying with $a=0$ and $b=3$ gives that $x^{1+1/(x+3)}-(x+\log x)\to 0$.

Subtracting one from the other gives $(x+3)^{1+1/x}-x^{1+1/(x+3)}\to 3$.

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