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

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