How to solve this limit using Taylor formula? I have difficulties in solving this limit using Taylor formula.
$$ \lim_{x \rightarrow+ \infty} \big((x+a)^{1+\frac {1}{x}} - x^{1+\frac{1}{x+a}}\big) $$
Can you give me some advice? I will be very thankful for your help.

Let $a = bx$, 
$(x+a)^{1+\frac {1}{x}} - x^{1+\frac{1}{x+a}}=((b+1)x)^{1+\frac{1}{x}}-x^{1+\frac{1}{(1+b)x}}$
fixing $x$, expand at $b=0$
$x^{1+\frac{1}{x}}(1+b+o(b^2))-(x^{1+\frac{1}{x}}-\log(x)x^{1+\frac{1}{x}}\frac{1}{x}b+o(b^2))=x^{1+\frac{1}{x}}(1+\frac{\log(x)}{x})b+o(b^2)=ax^{\frac{1}{x}}(1+\frac{\log(x)}{x})+o(b^2)$
This converges to $a$ as $x\rightarrow\infty, b\rightarrow0$
