Limit of the difference between two exponential functions 
Find the limit $$\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).$$

I did nothing by now.
 A: We can proceed as follows
\begin{align}
L &= \lim_{x \to \infty}(x + 3)^{1 + 1/x} - x^{1 + 1/(x + 3)}\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 1)\log(x + 3)}{x}\right) - \exp\left(\frac{(x + 4)\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\left\{\exp\left(\frac{(x + 1)\log(x + 3)}{x} - \frac{(x + 4)\log x}{x + 3}\right) - 1\right\}\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\cdot\frac{e^{t} - 1}{t}\cdot t \text{ (note that }t \to 0\text{ as }x \to \infty)\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\cdot t\notag\\
&= \lim_{x \to \infty}\exp\left(\frac{(x + 4)\log x}{x + 3}\right)\left(\frac{(x + 1)\log(x + 3)}{x} - \frac{(x + 4)\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}x\cdot x^{1/(x + 3)}\left(\log\frac{x + 3}{x} + \frac{\log(x + 3)}{x} - \frac{\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}x\left(\log\frac{x + 3}{x} + \frac{\log(x + 3)}{x} - \frac{\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\left(x\log\frac{x + 3}{x} + \log(x + 3) - \frac{x\log x}{x + 3}\right)\notag\\
&= \lim_{x \to \infty}\left(3\cdot\dfrac{\log(1 + 3/x)}{3/x} + \log\frac{x + 3}{x} + 3\cdot\frac{\log x}{x + 3}\right)\notag\\
&= 3\cdot 1 + \log 1 + 3\cdot 0\notag\\
&= 3\notag
\end{align}
Here we have used the standard limits $$\lim_{h \to 0}\frac{\log(1 + h)}{h} = 1,\,\lim_{x \to \infty}\frac{\log x}{x} = 0$$ and because of these limits the expression $$x^{1/(x + 3)} = \exp\left(\frac{\log x}{x + 3}\right) \to 1$$ as $x \to \infty$ and the variable $$t = \frac{(x + 1)\log(x + 3)}{x} - \frac{(x + 4)\log x}{x + 3} = \log\frac{x + 3}{x} + \frac{\log(x + 3)}{x} - \frac{\log x}{x + 3}$$ tends to $0$ as $x \to \infty$.
A: We have that
$$\lim_{x \to +\infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right)=
\lim_{x \to +\infty}\left((x+3)\exp\left[\frac{\ln(x+3)}{x}\right]-x\exp\left[\frac{\ln(x)}{x+3}\right]\right).
$$
Moreover
$$(x+3)\exp\left[\frac{\ln(x+3)}{x}\right]=(x+3)\cdot\left(1+\frac{\ln(x+3)}{x}+O(\ln^2(x)/x^2)\right)\\
=x+3+(x+3)\frac{\ln(x+3)}{x}+o(1)$$
and
$$x\exp\left[\frac{\ln(x)}{x+3}\right]=x\cdot\left(1+\frac{\ln(x)}{x+3}+O(\ln^2(x)/x^2)\right)\\
=x+\frac{x\ln(x)}{x+3}+o(1).
$$
Hence we obtain
$$\lim_{x \to +\infty}\left(x+3+(x+3)\frac{\ln(x+3)}{x}-x-\frac{x\ln(x)}{x+3}\right)
=\lim_{x \to +\infty}\left(3+(x+3)\frac{\ln(x)+\ln(1+3/x)}{x}-\frac{x\ln(x)}{x+3}\right)\\
=\lim_{x \to +\infty}\left(3+\ln(x)\cdot \left(\frac{x+3}{x}-\frac{x}{x+3}\right)\right)\\
=\lim_{x \to +\infty}\left(3+\ln(x)\cdot \left(\frac{6x+9}{x(x+3)}\right)\right)=3.$$
