Limit of $(x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$ when $x\to \infty$ Okay, this is the last limit I have to solve but it's not that easy ;)
$$\lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$$
 A: 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: 
$$\begin{eqnarray*}
(t^{-1}+3)^{1+t} &=&t^{-1}+\left( 3-\ln t\right) +\left( -\frac{3}{2}+\frac{1
}{2}\left( 3-\ln t\right) ^{2}\right) t+O\left( t^{2}\right)  \\
t^{-1-1/(t^{-1}+3)} &=&t^{-1}+\left( -\ln t\right) +\left( 3\ln t+\frac{1}{2}
\ln ^{2}t\right) t+O\left( t^{2}\right) 
\end{eqnarray*}$$
and
$$(t^{-1}+3)^{1+t}-t^{-1-1/(t^{-1}+3)}=3+\left( -\frac{3}{2}+\frac{1}{2}\left(
3-\ln t\right) ^{2}-3\ln t-\frac{1}{2}\ln ^{2}t\right) t+O\left(
t^{2}\right).$$
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
$$\begin{eqnarray*}
f(x)&=&(x+3)^{1+1/x}-x^{1+1/(x+3)} \\ &=&\frac{\left( x+3\right)
^{1+1/x}x^{-1-1/(x+3)}-1}{x^{-1-1/(x+3)}} \\
&\sim &\frac{1+\frac{3}{x}+\left( -3\ln \frac{1}{x}+3\right) \frac{1}{x^{2}}
+O\left( \frac{1}{x^{3}}\right) -1}{\frac{1}{x}+\left( \ln \frac{1}{x}
\right) \frac{1}{x^{2}}+O\left( \frac{1}{x^{3}}\right) } \\
&\sim &3\frac{x+\ln x+1}{x-\ln x} \\
&\sim &3.
\end{eqnarray*}$$
A: 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 $3$.
A: 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 $(x+3)^{1+1/x}-x^{1+1/(x+3)}\to 3$.
A: $(x+3)e^{((\ln(x+3))/x)}-xe^{((\ln x)/(x+3))}=
3+{(((x+3)\ln(x+3))/x)-((x\ln x)/(x+3))}+o(1)=
3+o(1)$, 
where the first equality is obtained by Maclaurin series of the exponential function and the last equality is obtained by the trick  $\ln(x+3)= \ln x+\ln(1+(3/x))$.
