<s>[INST] 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 ;)
$$\lim_{x\to \infty} (x+3)^{1 + 1/x} - x^{1 + 1/(x+3)}$$
 [/INST] 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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