How to calculate $\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$ How to calculate  $$\lim \limits_{x \to \infty} \left[(x+a)^{1+\frac1x}-(x) ^{1+\frac{1}{x+a}}\right]$$ 
The limit equals a but any hints for the method? 
 A: Let us consider $$A=(x+a)^{1+\frac1x}\qquad \text {and} \qquad B=(x) ^{1+\frac{1}{x+a}}$$
$$A=(x+a)^{1+\frac1x}\implies \log(A)=\left({1+\frac1x}\right)\log(x+a)$$ Using Taylor we then have
$$\log(A)=\log \left({x}\right)+\frac{a+\log
   \left({x}\right)}{x}+\frac{a-\frac{a^2}{2}}{x^2}+O\left(\frac{1}{x^3
   }\right)$$
$$A=e^{\log(A)}=x+a+\log \left({x}\right)+\frac{2
   \left(a-\frac{a^2}{2}\right)+\left(a+\log \left({x}\right)\right)^2}{2
   x}+O\left(\frac{1}{x^2}\right)$$
$$B=(x) ^{1+\frac{1}{x+a}}\implies \log(B)=\left({1+\frac{1}{x+a}}\right)\log(x)$$
$$\log(B)=\log \left({x}\right)+\frac{\log \left({x}\right)}{x}-\frac{a \log
   \left({x}\right)}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$B=e^{\log(B)}=x+\log \left({x}\right)+\frac{-2 a \log \left({x}\right)+\log
   ^2\left({x}\right)}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$A-B=a\left(1+\frac{1+2  \log
   \left({x}\right)}{x}\right)+O\left(\frac{1}{x^2}\right)$$
Trying with $x=10^6$ and $a=\pi$, the exact value would be $3.1416826014$ while the approximation leads to $3.1416826006$.
A: We can see two main things from a first glance at this problem. The first is that both exponents approach $1$ equivalently. Even though the denominator has $a$, as $x$ gets very large, the fraction contained in the exponent approaches $0$ at the same rate. Secondly, knowing that the exponents go to $1$, we compare $x+a$ to $x$. Note that we can only compare them since both exponents are equivalent in the sense of a limit. Clearly, $x+a-x$ is $a$, so the final answer is $a$.
