Behaviour of $f(x)=e-\left(1+\frac{1}{x} \right)^{x}$ when $x\to+\infty$ This is from an MCQ contest.

For all $x\geq 1$ let  $f(x)=e-\left(1+\dfrac{1}{x} \right)^{x}$ then we have :
  
  
*
  
*$f(x)\mathrel{\underset{_+\infty}{\sim}}\dfrac{e}{x}$ and $f$ is integrable on $[1,+\infty[$
  
*$f(x)\mathrel{\underset{_+\infty}{\sim}}\dfrac{e}{2\sqrt{x}}$ and $f$ is not integrable on $[1,+\infty[$
  
*$f(x)\mathrel{\underset{_+\infty}{\sim}}\dfrac{e}{\sqrt{2}x}$ and $f$ is integrable on $[1,+\infty[$
  
*$f(x)\mathrel{\underset{_+\infty}{\sim}}\dfrac{e}{2x}$ and $f$ is not integrable on $[1,+\infty[$
  

My thoughts :
i don't know how to prove that for example to show that a is true shall we procced like this 
$$\lim_{x\to +\infty} \dfrac{e-e^{ x\ln(1+\dfrac{1}{x}) } }{ \dfrac{e}{x}}=\lim_{x\to +\infty}x\left( 1-e^{x\ln(1+\dfrac{1}{x})}\right)$$

Answers 1. and 3. can be excluded a priori, because if $f(x)\sim k/x$ then $f$ cannot be integrable on $[1,+\infty[$.
Then it is quite obvious that we can make a Taylor expansion in $1/x$, which will never yield a $\sqrt{x}$ term: if so 2. can be excluded and the correct answer should be 4.
In fact, using $\ln(1+t)=t-t^2/2+O(t^3)$ one gets
$$
e-e^{x\ln(1+1/x)}=e - e^{1-1/(2x)+O(1/x^2)}=
e\left(1-e^{-1/(2x)+O(1/x^2)}\right)=e\left({1\over2x}+O\left({1\over x^2}\right)\right).
$$
